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Graphing Quadratic Equations in Two Variables

Recognize the graph of a quadratic equation in two variables

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form \(Ax+By=C\). We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form \(y=a{x}^{2}+bx+c\). We call this kind of equation a quadratic equation in two variables.

Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

Let’s look first at graphing the quadratic equation \(y={x}^{2}\). We will choose integer values of \(x\) between \(-2\) and 2 and find their \(y\) values. See .

\(y={x}^{2}\)
\(x\)	\(y\)
0	0
1	1
\(-1\)	1
2	4
\(-2\)	4

Notice when we let \(x=1\) and \(x=-1\), we got the same value for \(y\).

\[\begin{array}{llll}y={x}^{2} & & & y={x}^{2} \\ y={1}^{2} & & & y={(-1)}^{2} \\ y=1 & & & y=1\end{array}\]

The same thing happened when we let \(x=2\) and \(x=-2\).

Now, we will plot the points to show the graph of \(y={x}^{2}\). See .

Example

Try it.

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

Solution

We will graph the equation by plotting points.

Choose integers values for x, substitute them into the equation and solve for y.
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 equation\(y={x}^{2}-1\).

Example

Try it.

Determine whether each parabola opens upward or downward:

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

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.

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

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 that 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 in blue. 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 \(y=a{x}^{2}+bx+c\) 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 \(x=-\frac{b}{2a}\).

Look back at . Are these the equations of the dashed red lines?

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. 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 equation.

Example

Try it.

For the parabola \(y=3{x}^{2}-6x+2\) find: ⓐ the axis of symmetry and ⓑ the vertex.

Solution

ⓐ
The axis of symmetry is the line \(x=-\frac{b}{2a}\).
Substitute the values of a, b into the equation.
Simplify.	\(x=1\)
	The axis of symmetry is the line \(x=1\).
ⓑ
The vertex is on the line of symmetry, so its x-coordinate will be \(x=1\).
Substitute\(x=1\) into the equation and solve for y.
Simplify.
This is the y-coordinate.	\(y=-1\) The vertex is \((1,\text{-}1).\)

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 equation.

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

At an x-intercept, the value of \(y\) is zero. To find an x-intercept, we substitute \(y=0\) into the equation. In other words, we will need to solve the equation \(0=a{x}^{2}+bx+c\) for \(x\).

\[\begin{array}{l} \\ \\ y=a{x}^{2}+bx+c \\ 0=a{x}^{2}+bx+c\end{array}\]

But 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 shown in .

First, we will find the x-intercepts of a parabola with equation \(y={x}^{2}+4x+3\).


Let \(y=0\).
Factor.
Use the zero product property.
Solve.
	The x intercepts are \((\text{-}1,0)\) and \((\text{-}3,0).\)


Let \(y=0\).
This quadratic does not factor, so we use the Quadratic Formula.
\(a=-1\), \(b=4\), \(c=3\)
Simplify.
	The x intercepts are \((2+\sqrt{7},0)\) and \((2-\sqrt{7},0)\).

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

Example

Try it.

Find the intercepts of the parabola \(y={x}^{2}-2x-8\).

Solution


To find the y-intercept, let \(x=0\) and solve for y.
	When \(x=0\), then \(y=-8\). The y-intercept is the point \((0,-8)\).

To find the x-intercept, let \(y=0\) and solve for x.
Solve by factoring.

When \(y=0\), then \(x=4\ \text{or}\ x=-2\). The x-intercepts are the points \((4,0)\) and \((-2,0)\).

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

Graph Quadratic Equations in Two Variables

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

How To Graph a Quadratic Equation in Two Variables

Try it.

Graph \(y={x}^{2}-6x+8\).

Solution

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 \(y=\text{-}{x}^{2}+6x-9\).

Solution

The equation y has on one side.
Since a is \(-1\), the parabola opens downward. To find the axis of symmetry, find \(x=-\frac{b}{2a}\).		The axis of symmetry is \(x=3.\) The vertex is on the line \(x=3.\)
Find y when \(x=3.\)	The vertex is \((3,0).\)
The y-intercept occurs when \(x=0.\) Substitute \(x=0.\) Simplify. 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 y-intercept is \((0,-9).\)
The x-intercept occurs when \(y=0.\)
Substitute \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

For the graph of \(y=-{x}^{2}+6x-9\), 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 \(0=\text{-}{x}^{2}+6x-9\) 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 \(y={x}^{2}+4x+5\)?

Condensed — the full section is in OpenStax Elementary 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 equation. The y-coordinate of the vertex is the minimum y-value of a parabola that opens upward. It is the maximum y-value of a parabola that opens downward. See .

Example

Try it.

Find the minimum value of the quadratic equation \(y={x}^{2}+2x-8\).

Solution


Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the axis of symmetry.	The axis of symmetry is \(x=-1\).
The vertex is on the line \(x=-1.\)
Find y when \(x=-1.\)	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

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

to calculate the height in feet, \(h\), of an object shot upwards into the air with initial velocity, \({v}_{0}\), after \(t\) seconds.

This formula is a quadratic equation in the variable \(t\), so its graph is a parabola. By solving for the coordinates of the vertex, 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 Elementary Algebra 2e.

Key Concepts

The graph of every quadratic equation is a parabola.
Parabola Orientation For the quadratic equation \(y=a{x}^{2}+bx+c\), if
- \(a>0\), the parabola opens upward.
- \(a<0\), the parabola opens downward.
Axis of Symmetry and Vertex of a Parabola For a parabola with equation \(y=a{x}^{2}+bx+c\):
- The axis of symmetry of a parabola is the line \(x=-\frac{b}{2a}\).
- The vertex is on the axis of symmetry, so its x-coordinate is \(-\frac{b}{2a}\).
- To find the y-coordinate of the vertex we substitute \(x=-\frac{b}{2a}\) into the quadratic equation.
Find the Intercepts of a Parabola To find the intercepts of a parabola with equation \(y=a{x}^{2}+bx+c\):
\(\begin{array}{llll}\text{y}\text{-intercept} & & & \text{x}\text{-intercepts} \\ \text{Let}\ x=0\ \text{and solve for}\ y. & & & \text{Let}\ y=0\ \text{and solve for}\ x.\end{array}\)
To Graph a Quadratic Equation in Two Variables
1. Write the quadratic equation with \(y\) on one side.
2. Determine whether the parabola opens upward or downward.
3. Find the axis of symmetry.
4. Find the vertex.
5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
6. Find the x-intercepts.
7. 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.

Chapter 10 Review Exercises

In the following exercises, solve using the Square Root Property.

Try it.

\({x}^{2}=100\)

Solution

\(x=\pm \ 10\)

Try it.

\({y}^{2}=144\)

Try it.

\({m}^{2}-40=0\)

Solution

\(m=\pm \ 2\sqrt{10}\)

Try it.

\({n}^{2}-80=0\)

Try it.

\(4{a}^{2}=100\)

Solution

\(a=\pm \ 5\)

Try it.

\(2{b}^{2}=72\)

Try it.

\({r}^{2}+32=0\)

Solution

no solution

Try it.

\({t}^{2}+18=0\)

Try it.

\(\frac{4}{3}{v}^{2}+4=28\)

Solution

\(v=\pm \ 3\sqrt{2}\)

Try it.

\(\frac{2}{3}{w}^{2}-20=30\)

Try it.

\(5{c}^{2}+3=19\)

Solution

\(c=\pm \ \frac{4\sqrt{5}}{5}\)

Try it.

\(3{d}^{2}-6=43\)

In the following exercises, solve using the Square Root Property.

Try it.

\({(p-5)}^{2}+3=19\)

Solution

\(p=1,9\)

Try it.

\({(q+4)}^{2}=9\)

Try it.

\({(u+1)}^{2}=45\)

Solution

\(u=-1\pm 3\sqrt{5}\)

Try it.

\({(z-5)}^{2}=50\)

Try it.

\({(x-\frac{1}{4})}^{2}=\frac{3}{16}\)

Solution

\(x=\frac{1}{4}\pm \frac{\sqrt{3}}{4}\)

Try it.

\({(y-\frac{2}{3})}^{2}=\frac{2}{9}\)

Try it.

\({(m-7)}^{2}+6=30\)

Solution

\(m=7\pm 2\sqrt{6}\)

Try it.

\({(n-4)}^{2}-50=150\)

Try it.

\({(5c+3)}^{2}=-20\)

Solution

no solution

Try it.

\({(4c-1)}^{2}=-18\)

Try it.

\({m}^{2}-6m+9=48\)

Solution

\(m=3\pm 4\sqrt{3}\)

Try it.

\({n}^{2}+10n+25=12\)

Try it.

\(64{a}^{2}+48a+9=81\)

Solution

\(a=-\frac{3}{2},\frac{3}{4}\)

Try it.

\(4{b}^{2}-28b+49=25\)

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

Try it.

\({x}^{2}+22x\)

Solution

\({(x+11)}^{2}\)

Try it.

\({y}^{2}+6y\)

Try it.

\({m}^{2}-8m\)

Solution

\({(m-4)}^{2}\)

Try it.

\({n}^{2}-10n\)

Try it.

\({a}^{2}-3a\)

Solution

\({(a-\frac{3}{2})}^{2}\)

Try it.

\({b}^{2}+13b\)

Try it.

\({p}^{2}+\frac{4}{5}p\)

Solution

\({(p+\frac{2}{5})}^{2}\)

Try it.

\({q}^{2}-\frac{1}{3}q\)

In the following exercises, solve by completing the square.

Try it.

\({c}^{2}+20c=21\)

Solution

\(c=1,-21\)

Try it.

\({d}^{2}+14d=-13\)

Try it.

\({x}^{2}-4x=32\)

Solution

\(x=-4,8\)

Try it.

\({y}^{2}-16y=36\)

Try it.

\({r}^{2}+6r=-100\)

Solution

no solution

Try it.

\({t}^{2}-12t=-40\)

Try it.

\({v}^{2}-14v=-31\)

Solution

\(v=7\pm 3\sqrt{2}\)

Try it.

\({w}^{2}-20w=100\)

Try it.

\({m}^{2}+10m-4=-13\)

Solution

\(m=-9,-1\)

Try it.

\({n}^{2}-6n+11=34\)

Try it.

\({a}^{2}=3a+8\)

Solution

\(a=\frac{3}{2}\pm \frac{\sqrt{41}}{2}\)

Try it.

\({b}^{2}=11b-5\)

Try it.

\((u+8)(u+4)=14\)

Solution

\(u=-6\pm 3\sqrt{2}\)

Try it.

\((z-10)(z+2)=28\)

Try it.

\(3{p}^{2}-18p+15=15\)

Solution

\(p=0,6\)

Try it.

\(5{q}^{2}+70q+20=0\)

Try it.

\(4{y}^{2}-6y=4\)

Solution

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

Try it.

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

Try it.

\(3{c}^{2}+2c=9\)

Solution

\(c=-\frac{1}{3}\pm \frac{2\sqrt{7}}{3}\)

Try it.

\(4{d}^{2}-2d=8\)

Condensed — the full section is in OpenStax Elementary 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.

Graph the equation \(y=3x-5\) by plotting points.
If you missed this problem, review .

Жауап беріңіз
Evaluate \(2{x}^{2}+4x-1\) when \(x=-3\).
If you missed this problem, review .

Жауап беріңіз

\(5\)
Evaluate \(-\frac{b}{2a}\) when \(a=\frac{1}{3}\) and \(b=\frac{5}{6}\).
If you missed this problem, review .

Жауап беріңіз

\(-\frac{5}{4}\)

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

Жауап беріңіз

We will graph the equation by plotting points.

Choose integers values for x, substitute them into the equation and solve for y.
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 equation\(y={x}^{2}-1\).

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

Жауап беріңіз
Graph \(y={x}^{2}+1\).

Жауап беріңіз
Determine whether each parabola opens upward or downward:
ⓐ \(y=-3{x}^{2}+2x-4\) ⓑ \(y=6{x}^{2}+7x-9\)

Жауап беріңіз

ⓐ
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.
Determine whether each parabola opens upward or downward:
ⓐ \(y=2{x}^{2}+5x-2\) ⓑ \(y=-3{x}^{2}-4x+7\)

Жауап беріңіз

ⓐ up ⓑ down
Determine whether each parabola opens upward or downward:
ⓐ \(y=-2{x}^{2}-2x-3\) ⓑ \(y=5{x}^{2}-2x-1\)

Жауап беріңіз

ⓐ down ⓑ up

For the parabola \(y=3{x}^{2}-6x+2\) find: ⓐ the axis of symmetry and ⓑ the vertex.

Жауап беріңіз

ⓐ
The axis of symmetry is the line \(x=-\frac{b}{2a}\).
Substitute the values of a, b into the equation.
Simplify.	\(x=1\)
	The axis of symmetry is the line \(x=1\).
ⓑ
The vertex is on the line of symmetry, so its x-coordinate will be \(x=1\).
Substitute\(x=1\) into the equation and solve for y.
Simplify.
This is the y-coordinate.	\(y=-1\) The vertex is \((1,\text{-}1).\)

For the parabola \(y=2{x}^{2}-8x+1\) find: ⓐ the axis of symmetry and ⓑ the vertex.

Жауап беріңіз

ⓐ \(x=2\) ⓑ \((2,-7)\)
For the parabola \(y=2{x}^{2}-4x-3\) find: ⓐ the axis of symmetry and ⓑ the vertex.

Жауап беріңіз

ⓐ \(x=1\) ⓑ \((1,-5)\)
Find the intercepts of the parabola \(y={x}^{2}-2x-8\).

Жауап беріңіз

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

Жауап беріңіз

\(y\text{:}\ (0,-8);x\text{:}\ (-4,0),(2,0)\)
Find the intercepts of the parabola \(y={x}^{2}-4x-12.\)

Жауап беріңіз

\(y\text{:}\ (0,-12);x\text{:}\ (6,0),(-2,0)\)

Find the intercepts of the parabola \(y=5{x}^{2}+x+4\).

Жауап беріңіз


To find the y-intercept, let \(x=0\) and solve for y.	When \(x=0\), then \(y=4\). The y-intercept is the point \((0,4)\).

To find the x-intercept, let \(y=0\) and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.	\(\begin{array}{lllllll}\begin{array}{lll}{b}^{2} & - & 4ac \\ {1}^{2} & - & 4⋅5⋅4 \\ 1 & - & 80\end{array} \\ -79\ \end{array}\)
Since the value of the discriminant is negative, there is no real solution to the equation.	There are no x-intercepts.

Find the intercepts of the parabola \(y=3{x}^{2}+4x+4.\)

Жауап беріңіз

\(y\text{:}\ (0,4);x\text{:}\ \text{none}\)
Find the intercepts of the parabola \(y={x}^{2}-4x-5.\)

Жауап беріңіз

\(y\text{:}\ (0,-5);x\text{:}\ (5,0)\ (-1,0)\)

Find the intercepts of the parabola \(y=4{x}^{2}-12x+9\).

Жауап беріңіз


To find the y-intercept, let \(x=0\) and solve for y.
	When \(x=0\), then \(y=9\). The y-intercept is the point \((0,9)\).

To find the x-intercept, let \(y=0\) and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.	\(\begin{array}{lllllll}\begin{array}{lll}{b}^{2} & - & 4ac \\ {2}^{2} & - & 4⋅4⋅9 \\ 144 & - & 144\end{array} \\ 0\ \end{array}\)
	Since the value of the discriminant is 0, there is only one real solution to the equation. Therefore, there is only one x-intercept.
Solve the equation by factoring the perfect square trinomial.
Use the Zero Product Property.
Solve for x.
	When \(y=0\), then \(\frac{3}{2}=x.\)
	The x-intercept is the point \((\frac{3}{2},0).\)

Find the intercepts of the parabola \(y=\text{-}{x}^{2}-12x-36.\)

Жауап беріңіз

\(y\text{:}\ (0,-36);x\text{:}\ (-6,0)\)
Find the intercepts of the parabola \(y=9{x}^{2}+12x+4.\)

Жауап беріңіз

\(y\text{:}\ (0,4);x\text{:}\ (-\frac{2}{3},0)\)
Graph \(y={x}^{2}-6x+8\).
Graph the parabola \(y={x}^{2}+2x-8.\)

Жауап беріңіз

\(y\text{:}\ (0,-8)\); \(x\text{:}\ (2,0),(-4,0)\);
axis: \(x=-1\); vertex: \((-1,-9)\);
Graph the parabola \(y={x}^{2}-8x+12.\)

Жауап беріңіз

\(y\text{:}\ (0,12);\ x\text{:}\ (2,0),(6,0);\)
axis: \(x=4;\ \text{vertex:}\ (4,-4)\);

Graph \(y=\text{-}{x}^{2}+6x-9\).

Жауап беріңіз

The equation y has on one side.
Since a is \(-1\), the parabola opens downward. To find the axis of symmetry, find \(x=-\frac{b}{2a}\).		The axis of symmetry is \(x=3.\) The vertex is on the line \(x=3.\)
Find y when \(x=3.\)	The vertex is \((3,0).\)
The y-intercept occurs when \(x=0.\) Substitute \(x=0.\) Simplify. 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 y-intercept is \((0,-9).\)
The x-intercept occurs when \(y=0.\)
Substitute \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

Graph the parabola \(y=-3{x}^{2}+12x-12.\)

Жауап беріңіз

\(y\text{:}\ (0,-12);\ x\text{:}\ (2,0);\)
axis: \(x=2;\ \text{vertex:}(2,0)\);
Graph the parabola \(y=25{x}^{2}+10x+1.\)

Жауап беріңіз

\(y\text{:}\ (0,1);\ x\text{:}\ (-\frac{1}{5},0);\)
axis: \(x=-\frac{1}{5};\ \text{vertex:}\ (-\frac{1}{5},0)\);

Graph \(y={x}^{2}+4x+5\).

Жауап беріңіз

The equation has y on one side.
Since a is 1, the parabola opens upward.
To find the axis of symmetry, find \(x=-\frac{b}{2a}.\)	The axis of symmetry is \(x=-2.\)
The vertex is on the line \(x=-2.\)
Find y when \(x=-2.\)	The vertex is \((-2,1).\)
The y-intercept occurs when \(x=0.\) Substitute \(x=0.\) Simplify. The point \((0,5)\) is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is \((-4,5).\)	The y-intercept is \((0,5).\) Point symmetric to the y- intercept is \((-4,5)\).
The x- intercept occurs when \(y=0.\)
Substitute \(y=0.\) Test the discriminant.
		\({b}^{2}-4ac\) \({4}^{2}-4⋅15\) \(16-20\) \(\ -4\)
Since the value of the discriminant is negative, there is no solution and so no x- intercept. Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.

Graph the parabola \(y=2{x}^{2}-6x+5.\)

Жауап беріңіз

\(y\text{:}\ (0,5);\ x\text{:}\ \text{none};\)
axis: \(x=\frac{3}{2};\ \text{vertex:}\ (\frac{3}{2},\frac{1}{2})\);
Graph the parabola \(y=-2{x}^{2}-1.\)

Жауап беріңіз

\(y\text{:}\ (0,-1);x\text{:}\ \text{none};\)
axis: \(x=0;\ \text{vertex:}\ (0,-1)\);

Graph \(y=2{x}^{2}-4x-3\).

Жауап беріңіз


The equation y has one side. Since a is 2, the parabola opens upward.
To find the axis of symmetry, find \(x=-\frac{b}{2a}\).	The axis of symmetry is \(x=1\).
The vertex on the line \(x=1.\)
Find y when \(x=1\).	The vertex is \((1,\text{-}5)\).
The y-intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.	The y-intercept is \((0,-3)\).
The point \((0,-3)\) is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is \((2,-3)\)	Point symmetric to the y-intercept is \((2,-3).\)
The x-intercept occurs when \(y=0\).
Substitute \(y=0\).
Use the Quadratic Formula.
Substitute in the values of a, b, 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.

Graph the parabola \(y=5{x}^{2}+10x+3.\)

Жауап беріңіз

\(y\text{:}\ (0,3);\ x\text{:}\ (-1.6,0),(-0.4,0);\)
axis: \(x=-1;\ \text{vertex:}\ (-1,-2)\);
Graph the parabola \(y=-3{x}^{2}-6x+5.\)

Жауап беріңіз

\(y\text{:}\ (0,5);\ x\text{:}\ (0.6,0),(-2.6,0);\)
axis: \(x=-1;\ \text{vertex:}\ (-1,8)\);

Find the minimum value of the quadratic equation \(y={x}^{2}+2x-8\).

Жауап беріңіз


Since a is positive, the parabola opens upward.
The quadratic equation has a minimum.
Find the axis of symmetry.	The axis of symmetry is \(x=-1\).
The vertex is on the line \(x=-1.\)
Find y when \(x=-1.\)	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.

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

Жауап беріңіз

The minimum value is \(-4\) when \(x=4\).
Find the maximum or minimum value of the quadratic equation \(y=-4{x}^{2}+16x-11\).

Жауап беріңіз

The maximum value is 5 when \(x=2\).

The quadratic equation \(h=-16{t}^{2}+{v}_{0}t+{h}_{0}\) 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.

Жауап беріңіз

\(h=-16{t}^{2}+176t+4\)

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

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

ⓑ

Find h when \(t=5.5\).
Use a calculator to simplify.
	The vertex is \((5.5,488)\).
Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation.	The maximum value of the quadratic is 488 feet and it occurs when \(t=5.5\) seconds.

The quadratic equation \(h=-16{t}^{2}+128t+32\) 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? Round answers to the nearest tenth.

Жауап беріңіз

It will take 4 seconds to reach the maximum height of 288 feet.
A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of \(h=-16{t}^{2}+208t\). When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.

Жауап беріңіз

It will take 6.5 seconds to reach the maximum height of 676 feet.
\(y={x}^{2}+3\)

Жауап беріңіз

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.

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: Graphing Quadratic Equations in Two Variables

Recognize the graph of a quadratic equation in two variables
Find the axis of symmetry and vertex of a parabola
Find the intercepts of a parabola
Graph quadratic equations in two variables
Solve maximum and minimum applications
The axis of symmetry of a parabola is the line
The vertex is on the axis of symmetry, so its
Write the quadratic equation with

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.

Өзіңіздіңіңізді сынап көріңіз

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

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