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Parabolas

Graph vertical parabolas

Graph Vertical Parabolas

The next conic section we will look at is a parabola. We define a parabola as all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here. We will summarize the properties here.

Vertical Parabolas
	General form \(y=a{x}^{2}+bx+c\)	Standard form \(y=a{(x-h)}^{2}+k\)
Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
Axis of symmetry	\(x=-\frac{b}{2a}\)	\(x=h\)
Vertex	Substitute \(x=-\frac{b}{2a}\) and solve for y.	\((h,k)\)
y-intercept	Let \(x=0\)	Let \(x=0\)
x-intercepts	Let \(y=0\)	Let \(y=0\)

The graphs show what the parabolas look like when they open up or down. Their position in relation to the x- or y-axis is merely an example.

To graph a parabola from these forms, we used the following steps.

The next example reviews the method of graphing a parabola from the general form of its equation.

Example

Try it.

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

Solution


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 \(x=3.\)
Let \(x=3.\)


	The vertex is \((3,1).\)

The y-intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.
	The y-intercept is \((0,-8).\)
The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y-intercept is \((6,-8).\)

The x-intercept occurs when \(y=0.\)
Let \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x.
	The x-intercepts are \((4,0),(2,0).\)
Graph the parabola.

The next example reviews the method of graphing a parabola from the standard form of its equation, \(y=a{(x-h)}^{2}+k.\)

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

Graph Horizontal Parabolas

Our work so far has only dealt with parabolas that open up or down. We are now going to look at horizontal parabolas. These parabolas open either to the left or to the right. If we interchange the x and y in our previous equations for parabolas, we get the equations for the parabolas that open to the left or to the right.

Horizontal Parabolas
	General form \(x=a{y}^{2}+by+c\)	Standard form \(x=a{(y-k)}^{2}+h\)
Orientation	\(a>0\) right; \(a<0\) left	\(a>0\) right; \(a<0\) left
Axis of symmetry	\(y=-\frac{b}{2a}\)	\(y=k\)
Vertex	Substitute \(y=-\frac{b}{2a}\) and solve for x.	\((h,k)\)
y-intercepts	Let \(x=0\)	Let \(x=0\)
x-intercept	Let \(y=0\)	Let \(y=0\)

The graphs show what the parabolas look like when they to the left or to the right. Their position in relation to the x- or y-axis is merely an example.

Looking at these parabolas, do their graphs represent a function? Since both graphs would fail the vertical line test, they do not represent a function.

To graph a parabola that opens to the left or to the right is basically the same as what we did for parabolas that open up or down, with the reversal of the x and y variables.

Example

Try it.

Graph \(x=2{y}^{2}\) by using properties.

Solution


Since \(a=2,\) the parabola opens to the right.

To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)


	The axis of symmetry is \(y=0.\)
The vertex is on the line\(y=0.\)
Let \(y=0.\)

	The vertex is \((0,0).\)

Since the vertex is \((0,0),\) both the x- and y-intercepts are the point \((0,0).\) To graph the parabola we need more points. In this case it is easiest to choose values of y.

We also plot the points symmetric to \((2,1)\) and \((8,2)\) across the y-axis, the points \((2,-1),\)\((8,-2).\)

Graph the parabola.

In the next example, the vertex is not the origin.

In , we see the relationship between the equation in standard form and the properties of the parabola. The How To box lists the steps for graphing a parabola in the standard form \(x=a{(y-k)}^{2}+h.\) We will use this procedure in the next example.

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

Solve Applications with Parabolas

Many architectural designs incorporate parabolas. It is not uncommon for bridges to be constructed using parabolas as we will see in the next example.

Example

Try it.

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form\(y=a{(x-h)}^{2}+k.\)

Let the lower left side of the bridge be the origin of the coordinate grid at the point \((0,0).\) Since the base is 20 feet wide the point \((20,0)\) represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the y-coordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, \((0,0),\) and the rightmost point \((20,0).\) From this we know that the x-coordinate of the vertex will also be 10.
Identify the vertex, \((h,k).\)	\((h,k)=(10,10)\)
	\(h=10,\ k=10\)
Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola.	\(\ \begin{array}{lll}y & = & a{(x-h)}^{2}+k \\ y & = & a{(x-10)}^{2}+10 \\ (x,y) & = & (0,0)\end{array}\)
Substitute the values of the other point into the equation.	\(\ \begin{array}{lll}y & = & a{(x-10)}^{2}+10 \\ 0 & = & a{(0-10)}^{2}+10\end{array}\)
Solve for a.	\(\ \begin{array}{lll}0 & = & a{(0-10)}^{2}+10 \\ -10 & = & a{(-10)}^{2} \\ -10 & = & 100a \\ \frac{-10}{100} & = & a \\ a & = & -\frac{1}{10}\end{array}\)
	\(y=a{(x-10)}^{2}+10\)
Substitute the value for a into the equation.	\(\ y=-\frac{1}{10}{(x-10)}^{2}+10\)

Key Concepts

Parabola: A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Vertical Parabolas
	General form \(y=a{x}^{2}+bx+c\)	Standard form \(y=a{(x-h)}^{2}+k\)
Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
Axis of symmetry	\(x=-\frac{b}{2a}\)	\(x=h\)
Vertex	Substitute \(x=-\frac{b}{2a}\) and solve for y.	\((h,k)\)
y- intercept	Let \(x=0\)	Let \(x=0\)
x-intercepts	Let \(y=0\)	Let \(y=0\)

How to graph vertical parabolas \((y=a{x}^{2}+bx+c\) or \(f(x)=a{(x-h)}^{2}+k)\) using properties.

Determine whether the parabola opens upward or downward.
Find the axis of symmetry.
Find the vertex.
Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
Find the x-intercepts.
Graph the parabola.

Horizontal Parabolas
	General form \(x=a{y}^{2}+by+c\)	Standard form \(x=a{(y-k)}^{2}+h\)
Orientation	\(a>0\) right; \(a<0\) left	\(a>0\) right; \(a<0\) left
Axis of symmetry	\(y=-\frac{b}{2a}\)	\(y=k\)
Vertex	Substitute \(y=-\frac{b}{2a}\) and solve for x.	\((h,k)\)
y-intercepts	Let \(x=0\)	Let \(x=0\)
x-intercept	Let \(y=0\)	Let \(y=0\)

How to graph horizontal parabolas \((x=a{y}^{2}+by+c\) or \(x=a{(y-k)}^{2}+h)\) using properties.
1. Determine whether the parabola opens to the left or to the right.
2. Find the axis of symmetry.
3. Find the vertex.
4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
5. Find the y-intercepts.
6. Graph the parabola.

Parabolas

Graph Vertical Parabolas

In the following exercises, graph each equation by using properties.

Try it.

\(y=\text{-}{x}^{2}+4x-3\)

Solution

Try it.

\(y=\text{-}{x}^{2}+8x-15\)

Try it.

\(y=6{x}^{2}+2x-1\)

Solution

Try it.

\(y=8{x}^{2}-10x+3\)

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

Try it.

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

Solution

ⓐ \(y=\text{-}{(x-1)}^{2}-3\)
ⓑ

Try it.

\(y=2{x}^{2}+4x+6\)

Try it.

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

Solution

ⓐ \(y=-2{(x+1)}^{2}-3\)
ⓑ

Try it.

\(y=3{x}^{2}-12x+7\)

Graph Horizontal Parabolas

In the following exercises, graph each equation by using properties.

Try it.

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

Solution

Try it.

\(x=3{y}^{2}\)

Try it.

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

Solution

Try it.

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

Try it.

\(x=\text{-}{y}^{2}-2y+3\)

Solution

Try it.

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

Try it.

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

Solution

Try it.

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

Try it.

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

Solution

Try it.

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

Try it.

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

Solution

Try it.

\(x=\text{-}{(y-4)}^{2}+3\)

Try it.

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

Solution

Try it.

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

Try it.

\(x=\text{-}{(y+3)}^{2}+2\)

Solution

Try it.

\(x=\text{-}{(y+4)}^{2}+3\)

Try it.

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

Solution

Try it.

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

Try it.

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

Solution

Try it.

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

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

Try it.

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

Solution

ⓐ \(x={(y+2)}^{2}-9\)
ⓑ

Try it.

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

Try it.

\(x=-2{y}^{2}-12y-16\)

Solution

ⓐ \(x=-2{(y+3)}^{2}+2\)
ⓑ

Try it.

\(x=-3{y}^{2}-6y-5\)

Mixed Practice

In the following exercises, match each graph to one of the following equations: ⓐ x² + y² = 64 ⓑ x² + y² = 49
ⓒ (x + 5)² + (y + 2)² = 4 ⓓ (x − 2)² + (y − 3)² = 9 ⓔ y = −x² + 8x − 15 ⓕ y = 6x² + 2x − 1

Try it.

Solution

ⓐ

Try it.

Solution

ⓑ

Try it.

Solution

ⓓ

Try it.

Solve Applications with Parabolas

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

\(y=-\frac{1}{15}{(x-15)}^{2}+15\)

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

Solution

\(y=-\frac{1}{10}{(x-30)}^{2}+90\)

Try it.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

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.

Graph: \(y=-3{x}^{2}+12x-12.\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्
Solve by completing the square: \({x}^{2}-6x+6=0.\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\(x=3\pm \sqrt{3}\)
Write in standard form: \(y=3{x}^{2}-6x+5.\)
If you missed this problem, review .

जवाफ प्रकट गर्नुहोस्

\(y=3{\left(x-1\right)}^{2}+2\)

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

जवाफ प्रकट गर्नुहोस्


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 \(x=3.\)
Let \(x=3.\)


	The vertex is \((3,1).\)

The y-intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Simplify.
	The y-intercept is \((0,-8).\)
The point \((0,-8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,-8).\)	Point symmetric to the y-intercept is \((6,-8).\)

The x-intercept occurs when \(y=0.\)
Let \(y=0.\)
Factor the GCF.
Factor the trinomial.
Solve for x.
	The x-intercepts are \((4,0),(2,0).\)
Graph the parabola.

Graph \(y=\text{-}{x}^{2}+5x-6\) by using properties.

जवाफ प्रकट गर्नुहोस्
Graph \(y=\text{-}{x}^{2}+8x-12\) by using properties.

जवाफ प्रकट गर्नुहोस्

Write\(y=3{x}^{2}-6x+5\) in standard form and then use properties of standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्

Rewrite the function in \(y=a{(x-h)}^{2}+k\) form by completing the square.	\(y=3{x}^{2}-6x+5\)
	\(y=3({x}^{2}-2x\ )+5\)
	\(y=3({x}^{2}-2x+1)+5-3\)
	\(y=3{(x-1)}^{2}+2\)
Identify the constants a, h, k.	\(a=3\), \(h=1\), \(k=2\)
Since \(a=3,\) the parabola opens upward.

The axis of symmetry is \(x=h.\)	The axis of symmetry is \(x=1.\)
The vertex is \((h,k).\)	The vertex is \((1,2).\)
Find the y-intercept by substituting \(x=0.\)	\(y=3{(x-1)}^{2}+2\) \(y=3\cdot {0}^{2}-6\cdot 0+5\)
	\(y=5\)
	y-intercept \((0,5)\)
Find the point symmetric to \((0,5)\) across the axis of symmetry.	\((2,5)\)
Find the x-intercepts.	\(\begin{array}{lll}y & = & 3{(x-1)}^{2}+2 \\ 0 & = & 3{(x-1)}^{2}+2 \\ -2 & = & 3{(x-1)}^{2} \\ -\frac{2}{3} & = & {(x-1)}^{2} \\ \pm \sqrt{-\frac{2}{3}} & = & x-1\end{array}\)
	The square root of a negative number tells us the solutions are complex numbers. So there are no x-intercepts.
Graph the parabola.

ⓐ Write \(y=2{x}^{2}+4x+5\) in standard form and ⓑ use properties of standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्

ⓐ \(y=2{(x+1)}^{2}+3\)
ⓑ
ⓐ Write \(y=-2{x}^{2}+8x-7\) in standard form and ⓑ use properties of standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्

ⓐ \(y=-2{(x-2)}^{2}+1\)
ⓑ

Graph \(x=2{y}^{2}\) by using properties.

जवाफ प्रकट गर्नुहोस्


Since \(a=2,\) the parabola opens to the right.

To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)


	The axis of symmetry is \(y=0.\)
The vertex is on the line\(y=0.\)
Let \(y=0.\)

	The vertex is \((0,0).\)

Graph the parabola.

Graph \(x={y}^{2}\) by using properties.

जवाफ प्रकट गर्नुहोस्
Graph \(x=\text{-}{y}^{2}\) by using properties.

जवाफ प्रकट गर्नुहोस्

Graph \(x=\text{-}{y}^{2}+2y+8\) by using properties.

जवाफ प्रकट गर्नुहोस्


Since \(a=-1,\) the parabola opens to the left.

To find the axis of symmetry, find \(y=-\frac{b}{2a}.\)


	The axis of symmetry is \(y=1.\)
The vertex is on the line\(y=1.\)
Let \(y=1.\)

	The vertex is \((9,1).\)
The x-intercept occurs when \(y=0.\)


	The x-intercept is \((8,0).\)
The point \((8,0)\) is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is \((8,2)\)	Symmetric point is \((8,2).\)
The y-intercept occurs when \(x=0.\)
Substitute \(x=0.\)
Solve.


	The y-intercepts are \((0,4)\) and \((0,-2).\)
Connect the points to graph the parabola.

Graph \(x=\text{-}{y}^{2}-4y+12\) by using properties.

जवाफ प्रकट गर्नुहोस्
Graph \(x=\text{-}{y}^{2}+2y-3\) by using properties.

जवाफ प्रकट गर्नुहोस्

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

जवाफ प्रकट गर्नुहोस्


Identify the constants a, h, k.	\(a=2,\)\(h=1,\)\(k=2\)
Since \(a=2,\) the parabola opens to the right.

The axis of symmetry is \(y=k.\)	\(\\)The axis of symmetry is \(y=2.\)
The vertex is \((h,k).\)	\(\\)The vertex is \((1,2).\)
Find the x-intercept by substituting \(y=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ x & = & 2{(0-2)}^{2}+1 \\ x & = & 9\end{array}\)
	\(\\)The x-intercept is \((9,0).\)
Find the point symmetric to \((9,0)\) across the axis of symmetry.	\(\ (9,4)\)
Find the y-intercepts. Let \(x=0.\)	\(\begin{array}{lll}x & = & 2{(y-2)}^{2}+1 \\ 0 & = & 2{(y-2)}^{2}+1 \\ -1 & = & 2{(y-2)}^{2}\end{array}\)
	A square cannot be negative, so there is no real solution. So there are no y-intercepts.
Graph the parabola.

Graph \(x=3{(y-1)}^{2}+2\) using properties.

जवाफ प्रकट गर्नुहोस्
Graph \(x=2{(y-3)}^{2}+2\) using properties.

जवाफ प्रकट गर्नुहोस्

Graph \(x=-4{(y+1)}^{2}+4\) using properties.

जवाफ प्रकट गर्नुहोस्


Identify the constants a, h, k.	\(a=-4,\)\(h=4,\)\(k=-1\)
Since \(a=-4,\) the parabola opens to the left.

The axis of symmetry is \(y=k.\)	\(\\)The axis of symmetry is \(y=-1.\)
The vertex is \((h,k).\)	\(\\)The vertex is \((4,-1).\)
Find the x-intercept by substituting \(y=0.\)	\(\ \begin{array}{lll}x & = & -4{(y+1)}^{2}+4 \\ x & = & -4{(0+1)}^{2}+4 \\ x & = & 0\end{array}\)
	\(\\)The x-intercept is \((0,0).\)
Find the point symmetric to \((0,0)\) across the axis of symmetry.	\(\ (0,-2)\)
Find the y-intercepts.	\(\ x=-4{(y+1)}^{2}+4\)
Let \(x=0.\)	\(\begin{array}{lll}0 & = & -4{(y+1)}^{2}+4 \\ -4 & = & -4{(y+1)}^{2} \\ 1 & = & {(y+1)}^{2} \\ y+1 & = & \pm 1\end{array}\)
	\(y=-1+1\ y=-1-1\)
	\(y=0\ y=-2\)
	The y-intercepts are \((0,0)\) and \((0,-2).\)
Graph the parabola.

Graph \(x=-4{(y+2)}^{2}+4\) using properties.

जवाफ प्रकट गर्नुहोस्
Graph \(x=-2{(y+3)}^{2}+2\) using properties.

जवाफ प्रकट गर्नुहोस्

Write \(x=2{y}^{2}+12y+17\) in standard form and then use the properties of the standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्


Rewrite the function in \(x=a{(y-k)}^{2}+h\) form by completing the square.



Identify the constants a, h, k.	\(a=2,\ h=-1,\ k=-3\)
Since \(a=2,\) the parabola opens to the right.

The axis of symmetry is \(y=k.\)	\(\\)The axis of symmetry is \(y=-3.\)
The vertex is \((h,k).\)	\(\\)The vertex is \((-1,-3).\)
Find the x-intercept by substituting \(y=0.\)	\(\ \begin{array}{lll}x & = & 2{(y+3)}^{2}-1 \\ x & = & 2{(0+3)}^{2}-1 \\ x & = & 17\end{array}\)
	\(\\)The x-intercept is \((17,0).\)
Find the point symmetric to \((17,0)\) across the axis of symmetry.	\(\ (17,-6)\)
Find the y-intercepts. Let \(x=0.\)	\(\begin{array}{lll}x & = & 2{(y+3)}^{2}-1 \\ 0 & = & 2{(y+3)}^{2}-1 \\ 1 & = & 2{(y+3)}^{2} \\ \frac{1}{2} & = & {(y+3)}^{2} \\ y+3 & = & \pm \sqrt{\frac{1}{2}} \\ y & = & -3\pm \frac{\sqrt{2}}{2}\end{array}\)
	\(y=-3+\frac{\sqrt{2}}{2}\ y=-3-\frac{\sqrt{2}}{2}\)
	\(y\approx -2.3\ y\approx -3.7\)
	The y-intercepts are \((0,-3+\frac{\sqrt{2}}{2}),(0,-3-\frac{\sqrt{2}}{2}).\)
Graph the parabola.

ⓐ Write \(x=3{y}^{2}+6y+7\) in standard form and ⓑ use properties of the standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्

ⓐ \(x=3{(y+1)}^{2}+4\)
ⓑ
ⓐ Write \(x=-4{y}^{2}-16y-12\) in standard form and ⓑ use properties of the standard form to graph the equation.

जवाफ प्रकट गर्नुहोस्

ⓐ \(x=-4{(y+2)}^{2}+4\)
ⓑ

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

जवाफ प्रकट गर्नुहोस्

We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form\(y=a{(x-h)}^{2}+k.\)

Let the lower left side of the bridge be the origin of the coordinate grid at the point \((0,0).\) Since the base is 20 feet wide the point \((20,0)\) represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the y-coordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, \((0,0),\) and the rightmost point \((20,0).\) From this we know that the x-coordinate of the vertex will also be 10.
Identify the vertex, \((h,k).\)	\((h,k)=(10,10)\)
	\(h=10,\ k=10\)
Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola.	\(\ \begin{array}{lll}y & = & a{(x-h)}^{2}+k \\ y & = & a{(x-10)}^{2}+10 \\ (x,y) & = & (0,0)\end{array}\)
Substitute the values of the other point into the equation.	\(\ \begin{array}{lll}y & = & a{(x-10)}^{2}+10 \\ 0 & = & a{(0-10)}^{2}+10\end{array}\)
Solve for a.	\(\ \begin{array}{lll}0 & = & a{(0-10)}^{2}+10 \\ -10 & = & a{(-10)}^{2} \\ -10 & = & 100a \\ \frac{-10}{100} & = & a \\ a & = & -\frac{1}{10}\end{array}\)
	\(y=a{(x-10)}^{2}+10\)
Substitute the value for a into the equation.	\(\ y=-\frac{1}{10}{(x-10)}^{2}+10\)

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

जवाफ प्रकट गर्नुहोस्

\(y=-\frac{1}{20}{(x-20)}^{2}+20\)
Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

जवाफ प्रकट गर्नुहोस्

\(y=-\frac{1}{5}{(x-5)}^{2}+5\)
\(y=\text{-}{x}^{2}+4x-3\)

जवाफ प्रकट गर्नुहोस्
\(y=\text{-}{x}^{2}+8x-15\)
\(y=6{x}^{2}+2x-1\)

जवाफ प्रकट गर्नुहोस्
\(y=8{x}^{2}-10x+3\)
\(y=\text{-}{x}^{2}+2x-4\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(y=\text{-}{(x-1)}^{2}-3\)
ⓑ
\(y=2{x}^{2}+4x+6\)
\(y=-2{x}^{2}-4x-5\)

जवाफ प्रकट गर्नुहोस्

ⓐ \(y=-2{(x+1)}^{2}-3\)
ⓑ
\(y=3{x}^{2}-12x+7\)
\(x=-2{y}^{2}\)

जवाफ प्रकट गर्नुहोस्
\(x=3{y}^{2}\)
\(x=4{y}^{2}\)

जवाफ प्रकट गर्नुहोस्
\(x=-4{y}^{2}\)
\(x=\text{-}{y}^{2}-2y+3\)

जवाफ प्रकट गर्नुहोस्

Symbols used here

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|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: Parabolas

Graph vertical parabolas
Graph horizontal parabolas
Solve applications with parabolas
Determine whether the parabola opens upward or downward.
Find the axis of symmetry.
Find the vertex.
Find the
Find 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.

तपाईँको आफ्नै प्रयास गर्नुहोस्

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.

यसमा थप Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value