Unit 4, Lesson 13

Graphing the Standard Form (Part 2)

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Integrated Math 2

Practice

Problem 1

Here are four graphs. Match each graph with the quadratic equation that it represents.

<p>A curve in an x y plane, origin O.</p> — Graph A

Graph A
Graph B
Graph C
Graph D

$y = x^2+x$
$y = \text- x^2 +2$
$y = x^2 -x$
$y = x^2 + 3x$

Problem 2

Complete the table without graphing the equations.

equation	$x $-intercepts	$x $-coordinate of the vertex
$y=x^2+12x$
$y=x^2-3x$
$y=\text-x^2+16x$
$y=\text-x^2-24x$

Problem 3

Here is a graph that represents $y = x^2$.

Describe what would happen to the graph if the original equation were changed to $y=x^2-6x$. Predict the $x$- and $y$-intercepts of the graph and the quadrant where the vertex is located.

A curve in an x y plane, origin O. Horizontal axis, scale negative 8 to 8, by 2’s. Vertical axis, scale negative 12 to 12, by 2’s. A curve, labeled y equals x squared, passes through the points negative 2 comma 4, 0 comma 0, and 2 comma 4.
Sketch the graph of the equation $y=x^2 -6x$ on the same coordinate plane as $y=x^2$.

Problem 4

Select all equations whose graph opens upward.

$y=\text-x^2 + 9x$
$y=10x-5x^2$
$y=(2x-1)^2$
$y=(1-x)(2+x)$
$y=x^2-8x-7$

Problem 5

Technology required. Write an equation for a function that can be represented by each given graph. Then, use graphing technology to check each equation you wrote.

<p>Parabola. Opens up. X intercepts = 0 and -7. </p> — Graph 1

<p>Parabola. Opens up. Vertex = 0 comma -16. X intercepts = -4 and 4. </p> — Graph 2

<p>Parabola. Opens up. X intercepts = -9 and 3. </p> — Graph 3

Problem 6

ForLesson 8: Equivalent Quadratic Expressions

Match each quadratic expression that is written as a product with an equivalent expression that is expanded.

$(x+3)(x+4)$
$(x+3)(x+7)$
$(3x+4)(x+3)$
$(x+7)(3x+1)$

$x^2+10x+21$
$3x^2+13x+12$
$3x^2+22x+7$
$x^2+7x+12$

equation	\(x \)-intercepts	\(x \)-coordinate of the vertex
\(y=x^2+12x\)
\(y=x^2-3x\)
\(y=\text-x^2+16x\)
\(y=\text-x^2-24x\)

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