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Graph Quadratic Functions Using Properties

Recognize the graph of a quadratic function

Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function \(f(x)={x}^{2}\), which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form \(f(x)=a{x}^{2}+bx+c\) if \(a\ne 0.\) We call this kind of function a quadratic function.

We graphed the quadratic function \(f(x)={x}^{2}\) by plotting points.

Every quadratic function has a graph that looks like this. We call this figure a parabola.

Let’s practice graphing a parabola by plotting a few points.

Example

Try it.

Graph \(f(x)={x}^{2}-1.\)

Solution

We will graph the function by plotting points.

Choose integer values for x,
substitute them into the equation
and simplify to find \(f(x)\).

Record the values of the ordered pairs in the chart.
Plot the points, and then connect
them with a smooth curve. The
result will be the graph of the
function \(f(x)={x}^{2}-1\).

All graphs of quadratic functions of the form f (x) = ax2 + bx + c are parabolas that open upward or downward. See .

Notice that the only difference in the two functions is the negative sign before the quadratic term (x2 in the equation of the graph in ). When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

Example

Try it.

Determine whether each parabola opens upward or downward:

ⓐ \(f(x)=-3{x}^{2}+2x-4\) ⓑ \(f(x)=6{x}^{2}+7x-9.\)

Solution

ⓐ

Find the value of “a”.
Since the “a” is negative, the parabola will open downward.

ⓑ

Find the value of “a”.
Since the “a” is positive, the parabola will open upward.

Find the Axis of Symmetry and Vertex of a Parabola

Look again at . Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry. See .

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of f (x) = ax2 + bx + c is \(x=-\frac{b}{2a}.\)

So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula \(x=-\frac{b}{2a}.\)

Notice that these are the equations of the dashed blue lines on the graphs.

The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its
x-coordinate is \(-\frac{b}{2a}.\) To find the y-coordinate of the vertex we substitute the value of the x-coordinate into the quadratic function.

Example

Try it.

For the graph of \(f(x)=3{x}^{2}-6x+2\) find:

ⓐ the axis of symmetry ⓑ the vertex.

Solution

ⓐ

The axis of symmetry is the vertical line
\(x=-\frac{b}{2a}\).
Substitute the values of \(a,b\) into the
equation.
Simplify.
The axis of symmetry is the line \(x=1\).

ⓑ

The vertex is a point on the line of
symmetry, so its x-coordinate will be
\(x=1\).
Find \(f(1)\).
Simplify.
The result is the y-coordinate.
The vertex is \((1,-1)\).

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of x is zero. So to find the y-intercept, we substitute x = 0 into the function.

Let’s find the y-intercepts of the two parabolas shown in .

An x-intercept results when the value of f (x) is zero. To find an x-intercept, we let f (x) = 0. In other words, we will need to solve the equation 0 = ax2 + bx + c for x.

\[\begin{array}{lll}f(x) & = & a{x}^{2}+bx+c \\ 0 & = & a{x}^{2}+bx+c\end{array}\]

Solving quadratic equations like this is exactly what we have done earlier in this chapter!

We can now find the x-intercepts of the two parabolas we looked at. First we will find the x-intercepts of the parabola whose function is f (x) = x2 + 4x + 3.

Let \(f(x)=0\).
Factor.
Use the Zero Product Property.
Solve.
The x-intercepts are \((-1,0)\) and \((-3,0)\).

\[\begin{array}{llllll}(2+\sqrt{7},0)\approx (4.6,0) & & & & & (2-\sqrt{7},0)\approx (-0.6,0)\end{array}\]
Example

Try it.

Find the intercepts of the parabola whose function is \(f(x)={x}^{2}-2x-8.\)

Solution

To find the y-intercept, let \(x=0\) and
solve for \(f(x)\).
When \(x=0\), then \(f(0)=-8\).
The y-intercept is the point \((0,-8)\).
To find the x-intercept, let \(f(x)=0\) and
solve for \(x\).
Solve by factoring.
When \(f(x)=0\), then \(x=4\ \text{or}\ x=-2\).
The x-intercepts are the points \((4,0)\) and
\((-2,0)\).

\[\begin{array}{lllllllllllllllllllllllllll}\text{Quadratic equation} & & & & & & & & \ \text{Quadratic function} \\ \begin{array}{lll} \\ {x}^{2}-2x-15 & = & 0 \\ (x-5)(x+3) & = & 0 \\ x-5=0\ x+3 & = & 0 \\ x=5\ x & = & -3\end{array} & & & & \begin{array}{l}\text{Let}\ f(x)=0. \\ \\ \end{array} & & & & \ \begin{array}{lll}f(x) & = & {x}^{2}-2x-15 \\ 0 & = & {x}^{2}-2x-15 \\ 0 & = & (x-5)(x+3) \\ x-5 & = & 0\ x+3=0 \\ x & = & 5\ x=-3\end{array} \\ & & & & & & & & \ (5,0)\ \text{and}\ (-3,0) \\ & & & & & & & & \ x\text{-intercepts}\end{array}\]

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Graph Quadratic Functions Using Properties

Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.

How to Graph a Quadratic Function Using Properties

Try it.

Graph f (x) = x2 −6x + 8 by using its properties.

Solution

We list the steps to take in order to graph a quadratic function here.

We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.

Example

Try it.

Graph f (x) = –x2 + 6x − 9 by using its properties.

Solution

Since a is \(-1\), the parabola opens downward.
To find the equation of the axis of symmetry, use
\(x=-\frac{b}{2a}\).
The axis of symmetry is \(x=3\).
The vertex is on the line \(x=3\).
Find \(f(3)\).
The vertex is \((3,0).\)
The y-intercept occurs when \(x=0\). Find \(f(0)\).
Substitute \(x=0\).
Simplify.
The y-intercept is \((0,-9).\)
The point \((0,-9)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-9)\).
Point symmetric to the y-intercept is \((6,-9)\)
The x-intercept occurs when \(f(x)=0\).
Find \(f(x)=0\).
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

For the graph of f (x) = −x2 + 6x − 9, the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = −x2 + 6x − 9 is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.

How many x-intercepts would you expect to see on the graph of f (x) = x2 + 4x + 5?

Finding the y-intercept by finding f (0) is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the x-intercepts.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve Maximum and Minimum Applications

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The y-coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See .

Example

Try it.

Find the minimum or maximum value of the quadratic function \(f(x)={x}^{2}+2x-8.\)

Solution

Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the equation of the axis of symmetry.
The equation of the axis of
symmetry is \(x=-1\).
The vertex is on the line \(x=-1\).
Find \(f(-1)\).
The vertex is \((-1,-9)\).
Since the parabola has a minimum, the y-coordinate of
the vertex is the minimum y-value of the quadratic
equation.
The minimum value of the quadratic is \(-9\) and it
occurs when \(x=-1\).
Show the graph to verify the result.

We have used the formula

\[h(t)=-16{t}^{2}+{v}_{0}t+{h}_{0}\]

to calculate the height in feet, h , of an object shot upwards into the air with initial velocity, v0, after t seconds .

This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex (t, h), we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

  • Parabola Orientation
    • For the graph of the quadratic function \(f(x)=a{x}^{2}+bx+c,\) if
      • a > 0, the parabola opens upward.
      • a < 0, the parabola opens downward.
  • Axis of Symmetry and Vertex of a Parabola The graph of the function \(f(x)=a{x}^{2}+bx+c\) is a parabola where:
    • the axis of symmetry is the vertical line \(x=-\frac{b}{2a}.\)
    • the vertex is a point on the axis of symmetry, so its x-coordinate is \(-\frac{b}{2a}.\)
    • the y-coordinate of the vertex is found by substituting \(x=-\frac{b}{2a}\) into the quadratic equation.
  • Find the Intercepts of a Parabola
    • To find the intercepts of a parabola whose function is \(f(x)=a{x}^{2}+bx+c:\)
      \[\begin{array}{llllll}\text{y}\text{-intercept} & & & & & \text{x}\text{-intercepts} \\ \text{Let}\ x=0\ \text{and solve for}\ f(x). & & & & & \text{Let}\ f(x)=0\ \text{and solve for}\ x.\end{array}\]
  • How to graph a quadratic function using properties.
    1. Determine whether the parabola opens upward or downward.
    2. Find the equation of the axis of symmetry.
    3. Find the vertex.
    4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
    5. Find the x-intercepts. Find additional points if needed.
    6. Graph the parabola.
  • Minimum or Maximum Values of a Quadratic Equation
    • The y-coordinate of the vertex of the graph of a quadratic equation is the
    • minimum value of the quadratic equation if the parabola opens upward.
    • maximum value of the quadratic equation if the parabola opens downward.

Graph Quadratic Functions Using Properties

Recognize the Graph of a Quadratic Function

In the following exercises, graph the functions by plotting points.

Try it.

\(f(x)={x}^{2}+3\)

Solution

Try it.

\(f(x)={x}^{2}-3\)

Try it.

\(y=\text{-}{x}^{2}+1\)

Solution

Try it.

\(f(x)=\text{-}{x}^{2}-1\)

For each of the following exercises, determine if the parabola opens up or down.

Try it.

ⓐ \(f(x)=-2{x}^{2}-6x-7\) ⓑ \(f(x)=6{x}^{2}+2x+3\)

Solution

ⓐ down ⓑ up

Try it.

ⓐ \(f(x)=4{x}^{2}+x-4\) ⓑ \(f(x)=-9{x}^{2}-24x-16\)

Try it.

ⓐ \(f(x)=-3{x}^{2}+5x-1\) ⓑ \(f(x)=2{x}^{2}-4x+5\)

Solution

ⓐ down ⓑ up

Try it.

ⓐ \(f(x)={x}^{2}+3x-4\) ⓑ \(f(x)=-4{x}^{2}-12x-9\)

Find the Axis of Symmetry and Vertex of a Parabola

In the following functions, find ⓐ the equation of the axis of symmetry and ⓑ the vertex of its graph.

Try it.

\(f(x)={x}^{2}+8x-1\)

Solution

ⓐ \(x=-4\); ⓑ \((-4,-17)\)

Try it.

\(f(x)={x}^{2}+10x+25\)

Try it.

\(f(x)=\text{-}{x}^{2}+2x+5\)

Solution

ⓐ \(x=1\); ⓑ \((1,6)\)

Try it.

\(f(x)=-2{x}^{2}-8x-3\)

Find the Intercepts of a Parabola

In the following exercises, find the intercepts of the parabola whose function is given.

Try it.

\(f(x)={x}^{2}+7x+6\)

Solution

y-intercept: \((0,6);\) x-intercept \((-1,0),(-6,0)\)

Try it.

\(f(x)={x}^{2}+10x-11\)

Try it.

\(f(x)={x}^{2}+8x+12\)

Solution

y-intercept: \((0,12);\) x-intercept \((-2,0),(-6,0)\)

Try it.

\(f(x)={x}^{2}+5x+6\)

Try it.

\(f(x)=\text{-}{x}^{2}+8x-19\)

Solution

y-intercept: \((0,-19);\) x-intercept: none

Try it.

\(f(x)=-3{x}^{2}+x-1\)

Try it.

\(f(x)={x}^{2}+6x+13\)

Solution

y-intercept: \((0,13);\) x-intercept: none

Try it.

\(f(x)={x}^{2}+8x+12\)

Try it.

\(f(x)=4{x}^{2}-20x+25\)

Solution

y-intercept: \((0,25);\) x-intercept \((\frac{5}{2},0)\)

Try it.

\(f(x)=\text{-}{x}^{2}-14x-49\)

Try it.

\(f(x)=\text{-}{x}^{2}-6x-9\)

Solution

y-intercept: \((0,-9);\) x-intercept \((-3,0)\)

Try it.

\(f(x)=4{x}^{2}+4x+1\)

Graph Quadratic Functions Using Properties

In the following exercises, graph the function by using its properties.

Try it.

\(f(x)={x}^{2}+6x+5\)

Solution

Try it.

\(f(x)={x}^{2}+4x-12\)

Try it.

\(f(x)={x}^{2}+4x+3\)

Solution

Try it.

\(f(x)={x}^{2}-6x+8\)

Try it.

\(f(x)=9{x}^{2}+12x+4\)

Solution

Try it.

\(f(x)=\text{-}{x}^{2}+8x-16\)

Try it.

\(f(x)=\text{-}{x}^{2}+2x-7\)

Solution

Try it.

\(f(x)=5{x}^{2}+2\)

Try it.

\(f(x)=2{x}^{2}-4x+1\)

Solution

Try it.

\(f(x)=3{x}^{2}-6x-1\)

Try it.

\(f(x)=2{x}^{2}-4x+2\)

Solution

Try it.

\(f(x)=-4{x}^{2}-6x-2\)

Try it.

\(f(x)=\text{-}{x}^{2}-4x+2\)

Solution

Try it.

\(f(x)={x}^{2}+6x+8\)

Try it.

\(f(x)=5{x}^{2}-10x+8\)

Solution

Try it.

\(f(x)=-16{x}^{2}+24x-9\)

Try it.

\(f(x)=3{x}^{2}+18x+20\)

Solution

Try it.

\(f(x)=-2{x}^{2}+8x-10\)

Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value of each function.

Try it.

\(f(x)=2{x}^{2}+x-1\)

Solution

The minimum value is \(-\frac{9}{8}\) when \(x=-\frac{1}{4}.\)

Try it.

\(y=-4{x}^{2}+12x-5\)

Try it.

\(y={x}^{2}-6x+15\)

Solution

The minimum value is 6 when x = 3.

Try it.

\(y=\text{-}{x}^{2}+4x-5\)

Try it.

\(y=-9{x}^{2}+16\)

Solution

The maximum value is 16 when x = 0.

Try it.

\(y=4{x}^{2}-49\)

In the following exercises, solve. Round answers to the nearest tenth.

Try it.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic function h(t) = −16t2 + 168t + 45 find how long it will take the arrow to reach its maximum height, and then find the maximum height.

Solution

In 5.3 sec the arrow will reach maximum height of 486 ft.

Try it.

A stone is thrown vertically upward from a platform that is 20 feet height at a rate of 160 ft/sec. Use the quadratic function h(t) = −16t2 + 160t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.

Try it.

A ball is thrown vertically upward from the ground with an initial velocity of 109 ft/sec. Use the quadratic function h(t) = −16t2 + 109t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

Solution

In 3.4 seconds the ball will reach its maximum height of 185.6 feet.

Try it.

A ball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. Use the quadratic function h(t) = −16t2 + 122t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

Try it.

A computer store owner estimates that by charging x dollars each for a certain computer, he can sell 40 − x computers each week. The quadratic function R(x) = −x2 +40x is used to find the revenue, R, received when the selling price of a computer is x, Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

Solution

A selling price of $20 per computer will give the maximum revenue of $400.

Try it.

A retailer who sells backpacks estimates that by selling them for x dollars each, he will be able to sell 100 − x backpacks a month. The quadratic function R(x) = −x2 +100x is used to find the R, received when the selling price of a backpack is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

Try it.

A retailer who sells fashion boots estimates that by selling them for x dollars each, he will be able to sell 70 − x boots a week. Use the quadratic function R(x) = −x2 +70x to find the revenue received when the average selling price of a pair of fashion boots is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue per day.

Solution

A selling price of $35 per pair of boots will give a maximum revenue of $1,225.

Try it.

A cell phone company estimates that by charging x dollars each for a certain cell phone, they can sell 8 − x cell phones per day. Use the quadratic function R(x) = −x2 +8x to find the revenue received per day when the selling price of a cell phone is x. Find the selling price that will give them the maximum revenue per day, and then find the amount of the maximum revenue.

Try it.

A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation \(A(x)=x\left(120-\frac{x}{2}\right)\) gives the area of the corral, A, for the length, x, of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

Solution

The length of one side along the river is 120 feet and the maximum are is 7,200 square feet.

Try it.

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic function \(A(x)=x\left(50-\frac{x}{2}\right)\) gives the area, A, of the dog run for the length, x, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

Try it.

A land owner is planning to build a fenced in rectangular patio behind his garage, using his garage as one of the “walls.” He wants to maximize the area using 80 feet of fencing. The quadratic function A(x) = x(80 − 2x) gives the area of the patio, where x is the width of one side. Find the maximum area of the patio.

Solution

The maximum area of the patio is 800 feet.

Try it.

A family of three young children just moved into a house with a yard that is not fenced in. The previous owner gave them 300 feet of fencing to use to enclose part of their backyard. Use the quadratic function \(A(x)=x\left(150-\frac{x}{2}\right)\) determine the maximum area of the fenced in yard.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Graph the function \(f(x)={x}^{2}\) by plotting points.
    If you missed this problem, review .

    Zbulo përgjigjen

  2. Solve: \(2{x}^{2}+3x-2=0.\)
    If you missed this problem, review .

    Zbulo përgjigjen

    \(x=\frac{1}{2},\ x=-2\)

  3. Evaluate \(-\frac{b}{2a}\) when a = 3 and b = −6.
    If you missed this problem, review .

    Zbulo përgjigjen

    \(1\)

  4. Graph \(f(x)={x}^{2}-1.\)

    Zbulo përgjigjen

    We will graph the function by plotting points.

    Choose integer values for x,
    substitute them into the equation
    and simplify to find \(f(x)\).

    Record the values of the ordered pairs in the chart.
    Plot the points, and then connect
    them with a smooth curve. The
    result will be the graph of the
    function \(f(x)={x}^{2}-1\).

  5. Graph \(f(x)=\text{-}{x}^{2}.\).

    Zbulo përgjigjen

  6. Graph \(f(x)={x}^{2}+1.\)

    Zbulo përgjigjen

  7. Determine whether each parabola opens upward or downward:

    ⓐ \(f(x)=-3{x}^{2}+2x-4\) ⓑ \(f(x)=6{x}^{2}+7x-9.\)

    Zbulo përgjigjen

    ⓐ

    Find the value of “a”.
    Since the “a” is negative, the parabola will open downward.

    ⓑ

    Find the value of “a”.
    Since the “a” is positive, the parabola will open upward.

  8. Determine whether the graph of each function is a parabola that opens upward or downward:

    ⓐ \(f(x)=2{x}^{2}+5x-2\) ⓑ \(f(x)=-3{x}^{2}-4x+7.\)

    Zbulo përgjigjen

    ⓐ up; ⓑ down

  9. Determine whether the graph of each function is a parabola that opens upward or downward:

    ⓐ \(f(x)=-2{x}^{2}-2x-3\) ⓑ \(f(x)=5{x}^{2}-2x-1.\)

    Zbulo përgjigjen

    ⓐ down; ⓑ up

  10. For the graph of \(f(x)=3{x}^{2}-6x+2\) find:

    ⓐ the axis of symmetry ⓑ the vertex.

    Zbulo përgjigjen

    ⓐ

    The axis of symmetry is the vertical line
    \(x=-\frac{b}{2a}\).
    Substitute the values of \(a,b\) into the
    equation.
    Simplify.
    The axis of symmetry is the line \(x=1\).

    ⓑ

    The vertex is a point on the line of
    symmetry, so its x-coordinate will be
    \(x=1\).
    Find \(f(1)\).
    Simplify.
    The result is the y-coordinate.
    The vertex is \((1,-1)\).

  11. For the graph of \(f(x)=2{x}^{2}-8x+1\) find:

    ⓐ the axis of symmetry ⓑ the vertex.

    Zbulo përgjigjen

    ⓐ \(x=2;\) ⓑ \((2,-7)\)

  12. For the graph of \(f(x)=2{x}^{2}-4x-3\) find:

    ⓐ the axis of symmetry ⓑ the vertex.

    Zbulo përgjigjen

    ⓐ \(x=1;\) ⓑ \((1,-5)\)

  13. Find the intercepts of the parabola whose function is \(f(x)={x}^{2}-2x-8.\)

    Zbulo përgjigjen

    To find the y-intercept, let \(x=0\) and
    solve for \(f(x)\).
    When \(x=0\), then \(f(0)=-8\).
    The y-intercept is the point \((0,-8)\).
    To find the x-intercept, let \(f(x)=0\) and
    solve for \(x\).
    Solve by factoring.
    When \(f(x)=0\), then \(x=4\ \text{or}\ x=-2\).
    The x-intercepts are the points \((4,0)\) and
    \((-2,0)\).

  14. Find the intercepts of the parabola whose function is \(f(x)={x}^{2}+2x-8.\)

    Zbulo përgjigjen

    y-intercept: \((0,-8)\) x-intercepts \((-4,0),(2,0)\)

  15. Find the intercepts of the parabola whose function is \(f(x)={x}^{2}-4x-12.\)

    Zbulo përgjigjen

    y-intercept: \((0,-12)\) x-intercepts \((-2,0),(6,0)\)

  16. Find the intercepts of the parabola for the function \(f(x)=5{x}^{2}+x+4.\)

    Zbulo përgjigjen

    To find the y-intercept, let \(x=0\) and
    solve for \(f(x)\).
    When \(x=0\), then \(f(0)=4\).
    The y-intercept is the point \((0,4)\).
    To find the x-intercept, let \(f(x)=0\) and
    solve for \(x\).
    Find the value of the discriminant to
    predict the number of solutions which is
    also the number of x-intercepts.
    \(\begin{array}{l}{b}^{2}-4ac \\ {1}^{2}-4\cdot 5\cdot 4 \\ 1-80 \\ -79\end{array}\)
    Since the value of the discriminant is
    negative, there is no real solution to the
    equation.
    There are no x-intercepts.

  17. Find the intercepts of the parabola whose function is \(f(x)=3{x}^{2}+4x+4.\)

    Zbulo përgjigjen

    y-intercept: \((0,4)\) no x-intercept

  18. Find the intercepts of the parabola whose function is \(f(x)={x}^{2}-4x-5.\)

    Zbulo përgjigjen

    y-intercept: \((0,-5)\) x-intercepts \((-1,0),(5,0)\)

  19. Graph f (x) = x2 −6x + 8 by using its properties.

    Zbulo përgjigjen

  20. Graph f (x) = x2 + 2x − 8 by using its properties.

    Zbulo përgjigjen

  21. Graph f (x) = x2 − 8x + 12 by using its properties.

    Zbulo përgjigjen

  22. Graph f (x) = –x2 + 6x − 9 by using its properties.

    Zbulo përgjigjen

    Since a is \(-1\), the parabola opens downward.
    To find the equation of the axis of symmetry, use
    \(x=-\frac{b}{2a}\).
    The axis of symmetry is \(x=3\).
    The vertex is on the line \(x=3\).
    Find \(f(3)\).
    The vertex is \((3,0).\)
    The y-intercept occurs when \(x=0\). Find \(f(0)\).
    Substitute \(x=0\).
    Simplify.
    The y-intercept is \((0,-9).\)
    The point \((0,-9)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-9)\).
    Point symmetric to the y-intercept is \((6,-9)\)
    The x-intercept occurs when \(f(x)=0\).
    Find \(f(x)=0\).
    Factor the GCF.
    Factor the trinomial.
    Solve for x.
    Connect the points to graph the parabola.

  23. Graph f (x) = −3x2 + 12x − 12 by using its properties.

    Zbulo përgjigjen

  24. Graph f (x) = 4x2 + 24x + 36 by using its properties.

    Zbulo përgjigjen

  25. Graph f (x) = x2 + 4x + 5 by using its properties.

    Zbulo përgjigjen

    Since a is 1, the parabola opens upward.
    To find the axis of symmetry, find \(x=-\frac{b}{2a}\).
    The equation of the axis of symmetry is \(x=-2\).
    The vertex is on the line \(x=-2.\)
    Find \(f(x)\) when \(x=-2.\)
    The vertex is \((-2,1)\).
    The y-intercept occurs when \(x=0\).
    Find \(f(0).\)
    Simplify.
    The y-intercept is \((0,5)\).
    The point \((-4,5)\) is two units to the left of the line of
    symmetry.
    The point two units to the right of the line of
    symmetry is \((0,5)\).
    Point symmetric to the y-intercept is \((-4,5)\).
    The x-intercept occurs when \(f(x)=0\).
    Find \(f(x)=0\).
    Test the discriminant.
    Since the value of the discriminant is negative, there is
    no real solution and so no x-intercept.
    Connect the points to graph the parabola. You may
    want to choose two more points for greater accuracy.

  26. Graph f (x) = x2 − 2x + 3 by using its properties.

    Zbulo përgjigjen


  27. Graph f (x) = −3x2 − 6x − 4 by using its properties.

    Zbulo përgjigjen


  28. Graph f (x) = 2x2 − 4x − 3 by using its properties.

    Zbulo përgjigjen

    Since a is 2, the parabola opens upward.
    To find the equation of the axis of symmetry, use
    \(x=-\frac{b}{2a}\).
    The equation of the axis of
    symmetry is \(x=1.\)
    The vertex is on the line \(x=1\).
    Find \(f(1)\).
    The vertex is \((1,-5).\)
    The y-intercept occurs when \(x=0\).
    Find \(f(0)\).
    Simplify.
    The y-intercept is \((0,-3).\)
    The point \((0,-3)\) is one unit to the left of the line of
    symmetry.    		Point symmetric to the
    y-intercept is \((2,-3)\)
    The point one unit to the right of the line of
    symmetry is \((2,-3)\).
    The x-intercept occurs when \(y=0\).
    Find \(f(x)=0\).
    Use the Quadratic Formula.
    Substitute in the values of \(a,b,\) and \(c.\)
    Simplify.
    Simplify inside the radical.
    Simplify the radical.
    Factor the GCF.
    Remove common factors.
    Write as two equations.
    Approximate the values.
    The approximate values of the
    x-intercepts are \((2.5,0)\) and
    \((-0.6,0).\)
    Graph the parabola using the points found.

  29. Graph f (x) = 5x2 + 10x + 3 by using its properties.

    Zbulo përgjigjen

  30. Graph f (x) = −3x2 − 6x + 5 by using its properties.

    Zbulo përgjigjen

  31. Find the minimum or maximum value of the quadratic function \(f(x)={x}^{2}+2x-8.\)

    Zbulo përgjigjen

    Since a is positive, the parabola opens upward.
    The quadratic equation has a minimum.
    Find the equation of the axis of symmetry.
    The equation of the axis of
    symmetry is \(x=-1\).
    The vertex is on the line \(x=-1\).
    Find \(f(-1)\).
    The vertex is \((-1,-9)\).
    Since the parabola has a minimum, the y-coordinate of
    the vertex is the minimum y-value of the quadratic
    equation.
    The minimum value of the quadratic is \(-9\) and it
    occurs when \(x=-1\).
    Show the graph to verify the result.

  32. Find the maximum or minimum value of the quadratic function \(f(x)={x}^{2}-8x+12.\)

    Zbulo përgjigjen

    The minimum value of the quadratic function is −4 and it occurs when x = 4.

  33. Find the maximum or minimum value of the quadratic function \(f(x)=-4{x}^{2}+16x-11.\)

    Zbulo përgjigjen

    The maximum value of the quadratic function is 5 and it occurs when x = 2.

  34. The quadratic equation h(t) = −16t2 + 176t + 4 models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

    ⓐ How many seconds will it take the volleyball to reach its maximum height? ⓑ Find the maximum height of the volleyball.

    Zbulo përgjigjen

    \(\begin{array}{l}h(t)=-16{t}^{2}+176t+4\end{array}\)
    Since a is negative, the parabola opens downward.
    The quadratic function has a maximum.

    ⓐ
    Find the equation of the axis of symmetry.\(\begin{array}{lll} \\ \\ t & = & -\frac{b}{2a} \\ t & = & -\frac{176}{2(-16)} \\ t & = & 5.5\end{array}\)
    The equation of the axis of symmetry is \(t=5.5.\)
    The vertex is on the line \(t=5.5.\)The maximum occurs when \(t=5.5\) seconds.

    ⓑ
    Find \(h(5.5).\)



    Use a calculator to simplify.    		\(\begin{array}{lll} \\ h(t) & = & -16{t}^{2}+176t+4 \\ h(t) & = & -16{(5.5)}^{2}+176(5.5)+4 \\ h(t) & = & 488\end{array}\)
    The vertex is \((5.5,488).\)

    Since the parabola has a maximum, the h-coordinate of the vertex is the maximum value of the quadratic function.

    The maximum value of the quadratic is 488 feet and it occurs when t = 5.5 seconds.

    After 5.5 seconds, the volleyball will reach its maximum height of 488 feet.

  35. Solve, rounding answers to the nearest tenth.

    The quadratic function h(t) = −16t2 + 128t + 32 is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?

    Zbulo përgjigjen

    It will take 4 seconds for the stone to reach its maximum height of 288 feet.

  36. A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function of h(t) = −16t2 + 208t. When will the rocket reach its maximum height? What will be the maximum height?

    Zbulo përgjigjen

    It will 6.5 seconds for the rocket to reach its maximum height of 676 feet.

  37. \(f(x)={x}^{2}+3\)

    Zbulo përgjigjen

  38. \(f(x)={x}^{2}-3\)

  39. \(y=\text{-}{x}^{2}+1\)

    Zbulo përgjigjen

  40. \(f(x)=\text{-}{x}^{2}-1\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\approx
approximately equal
Equal to the precision shown, not exactly.
\neq
not equal
The two sides are different.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
|x|
absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i
imaginary unit
i² = −1.
\log_b x,\ \ln x
logarithm, natural log
The exponent b must be raised to for x; ln uses base e.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Graph Quadratic Functions Using Properties

  1. Recognize the graph of a quadratic function
  2. Find the axis of symmetry and vertex of a parabola
  3. Find the intercepts of a parabola
  4. Graph quadratic functions using properties
  5. Solve maximum and minimum applications
  6. the axis of symmetry is the vertical line
  7. the vertex is a point on the axis of symmetry, so its
  8. the

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Provo timen.

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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