﻿ Solve Quadratic Equation by Graphing Worksheet | Problems & Solutions

# Solve Quadratic Equation by Graphing Worksheet

Solve Quadratic Equation by Graphing Worksheet
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1.
The equation of the parabola A is $y$ = - $x$2. Find the equation of the parabola B that is obtained by the transformation of A.

 a. $y$ = - $x$ 2 + 3 b. $y$ = - $x$ 2 - 4 c. $y$ = - $x$ 2 + 4 d. $y$ = - $x$ 2 + 2

#### Solution:

The equation of the parabola A is = y = - x2.

The parabola A has been transformed 4 units up to get the parabola B.

That is, y = - (x 2) + 4.

So, the equation of the red parabola is y = - x 2 + 4.

2.
Identify the equation that transforms the parabola $A$, $y$ = 5$x$2 + 2 to the parabola $B$.

 a. $y$ = 5$x$2 - 4 b. $y$ = 5$x$2 - 1 c. $y$ = 5$x$2 + 4 d. $y$ = 5$x$2 + 1

#### Solution:

The equation of the parabola A is y = 5x2 + 2.

The parabola A has been transformed 6 units down to get the parabola B.

that is, y = (5x2 + 2) - 6 = 5x2 - 4.

So, the equation of the parabola B is y = 5x2 - 4.

3.
The graph of $y$ = 3$x$2 - 9$x$ is shown. Estimate the solution of 3$x$2 - 9$x$ = 0 using the graph.

 a. 4 and 8 b. 0 and 3 c. 0 and - 9 d. None of the above

#### Solution:

The x-intercepts in the graph of the equation y = ax2 + bx + c are the solutions of the related equation ax2 + bx + c = 0.

The graph intersects the x-axis at (0, 0) and (3, 0).

So, 0 and 3 are the solutions of the equation.

4.
Estimate the solution of $x$2 - $x$ = 6 using the graph.

 a. - 2 and 3 b. 12 and 30 c. 2 and 10 d. 2 and 3

#### Solution:

From the figure, the graph intersects the x-axis at two points (- 2,0) and (3,0).

The x-intercepts of the curve are - 2 and 3.
[x-coordinate of the point where curve meets x-axis.]

x2 - x = 6
[Original equation.]

(-2)2 - (- 2) = 6
[Substitute the x intercept = - 2.]

6 = 6
[Simplify.]

(3)2 - (3) = 6
[Substitute the x intercept = 3.]

6 = 6
[Simplify.]

Both the values satisfy the equation. So, -2 and 3 are the solutions of the equation.

5.
Estimate the solution of 2$x$2 - $x$ = 3 using the graph.

 a. -1 and $\frac{3}{2}$ b. 2 and 8 c. 1 and -2 d. -1 and 3

#### Solution:

From the figure, the graph intersects the x - axis at two points (-1, 0) and ( 3 / 2, 0).

The x-intercepts of the curve are -1 and 3 / 2.

2x2 - x = 3
[Original equation]

2(-1)2 - (-1) = 3
[Substitute x-intercept as -1.]

3 = 3
[Simplify.]

2( 32)2 - (32) = 3
[Substitute x-intercept as 3 / 2.]

3 = 3
[Simplify.]

Both the values satisfy the equation. So, -1 and 3 / 2 are the solutions of the equation.

6.
Find the solution of - 3$x$2 = - 27 using a graph.
 a. - 3 and 7 b. - 3 and 3 c. - 3 and 6 d. - 4 and 4

#### Solution:

- 3x2 = - 27
[Original equation]

x2 = 9
[Divide - 3 on each side]

x2 - 9 = 9 - 9
[Subtract 9 from each side]

x2 - 9 = 0
[Simplify.]

Sketch the graph of the related quadratic function y = x2 - 9 as shown below.

From the graph, the x-intercepts appear to be - 3 and 3.
[Estimate the values of the x-intercepts.]

So, - 3 and 3 are the solutions of the equation.

7.
Using a graph, estimate the solution of - $x$2 + 9$x$ - 20 = 0.
 a. 4 and -5 b. 4 and 8 c. 4 and 5 d. 1 and 5

#### Solution:

- x2 + 9x - 20 = 0 can be written in standard form as y = - x2 + 9x - 20.

Sketch the graph of the quadratic function y = - x2 + 9x - 20 as:

From the graph, the x-intercepts are 4 and 5.
[Estimate the values of the x-intercepts.]

- (4)2 + 9(4) - 20 = 0
[Substitute the x-intetcept = 4 in the equation - x2 - 20 = 0.]

0 = 0
[Simplify.]

- (5)2 + 9(5) - 20 = 0
[Substitute the x-intetcept = 5 in the equation - x2 - 20 = 0.]

0 = 0
[Simplify.]

Both the values satisfy the equation. So, 4 and 5 are the solutions of the equation.

8.
Find the roots of the quadratic equation $x$2 + 5 = 6 using the graph.

 a. -1 and 1 b. -1 and 4 c. -1 and 3 d. -3 and 1

#### Solution:

The equation in the standard form is ax2 + bx + c = 0.

x2 + 5 = 6
[Original equation.]

x2 + 5 - 6 = 6 - 6
[Subtract 6 from each side.]

x2 - 1 = 0
[Simplify.]

The graph appears to intersect the x-axis at the points (-1, 0) and (1, 0).

From the graph, the x-intercepts appear to be -1 and 1.
[Estimate the values of the x-intercepts.]

By substituting x = -1 and x = 1 in x2 - 1 = 0, -1 and 1 are obtained.

So, -1 and 1 are the roots of the quadratic equation.

9.
Using the graph, find the solution of -2$x$2 + 8 = 0.

 a. -5 and 3 b. -4 and 2 c. -2 and 2 d. -2 and 4

#### Solution:

-2x2 + 8 = 0 is written in the standard form as y = -2x2 + 8.

The graph intersect the x-axis at points (-2, 0) and (2, 0).

From the graph, the x-intercepts are -2 and 2.
[Estimate the values of the x-intercepts.]

-2(-2)2 + 8 = 0
[Substitute x = -2 in the equation -2x2 + 8 = 0.]

0 = 0
[Simplify.]

-2(2)2 + 8 = 0
[Substitute x = 2 in the equation -2x2 + 8 = 0.]

0 = 0
[Simplify.]

Both the values satisfy the equation. So, -2 and 2 are the solutions of the equation.

10.
Using a graph, identify the roots of the quadratic equation - $x$2 - 2$x$ + 3 = 0.
 a. 3 and 1 b. - 7 and 6 c. - 3 and 1 d. - 3 and 3

#### Solution:

- x2 - 2x + 3 = 0 is written in standard form as y = - x2 - 2x + 3.

Sketch the graph of the quadratic function y = - x2 - 2x + 3 as:

From the graph, the x-intercepts are - 3 and 1.
[Estimate the values of the x-intercepts.]

- (- 3)2 - 2(- 3) + 3 = 0
[Substitute x = - 3 in the equation - x2 - 2x + 3.]

0 = 0
[Simplify.]

- (1)2 - 2(1) + 3 = 0
[Substitute x = 1 in the equation - x2 - 2x + 3.]

0 = 0
[Simplify.]

Both the values satisfy the equation. So, - 3 and 1 are the roots of the equation.
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