The graph shows the height , in feet, of the top of an arch above the ground at horizontal distance , in feet. Which choice gives the width of the arch’s opening at ground level, in feet?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 88 | Free SAT Prep | Aniko

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Question 88·Easy·Nonlinear Functions

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The graph shows the height $y$ , in feet, of the top of an arch above the ground at horizontal distance $x$ , in feet.

Which choice gives the width of the arch’s opening at ground level, in feet?

When a quadratic graph models a height above the ground, the points on the ground are the $x$ -intercepts (where $y=0$ ). To find a ground-level width, read the two intercepts and subtract: (right intercept) $-$ (left intercept).

Hints

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Find the ground-level points

Ground level means the height is $0$ , so look for where the graph crosses the $x$ -axis.

Use the intercepts

Read the two $x$ -values where $y=0$ .

Compute the width

The width is the distance between those two $x$ -values (larger minus smaller).

Desmos Guide

Enter an equation with the same zeros

In Desmos, type $y=a(x-4)(x-16)$ so the graph will cross the $x$ -axis at $x=4$ and $x=16$ .

Adjust the shape to match the graph

Move the slider $a$ until the curve matches the displayed arch (it should open downward, so $a<0$ ).

Compute the width from the zeros

Use the two zeros $x=4$ and $x=16$ and compute the distance $16-4$ to get the width, which is $12$ feet.

Step-by-step Explanation

Identify where the arch meets the ground

The opening at ground level occurs where the height is $y=0$ . On the graph, these points are the $x$ -intercepts: $x=4$ and $x=16$ .

Find the distance between the intercepts

The width is the distance from $x=4$ to $x=16$ , which is $16-4=12$ . Therefore, the width of the opening is $12$ feet.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation