SAT Nonlinear Functions Question 164 | Free SAT Prep | Aniko

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Question 164·Medium·Nonlinear Functions

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An architect models the height of an arch with a quadratic function $h(x)$ , where $x$ is the horizontal distance (in feet) from the left end of the arch.

The arch passes through the points $(0,12)$ and $(6,12)$ , and its highest point is at $(3,21)$ .

Which choice is the value of $h(1)$ ?

When a quadratic’s highest or lowest point is given, use vertex form $a(x-h)^2+k$ immediately. Then plug in one additional point to find $a$ , and substitute the requested $x$ -value to get the function value.

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Use the vertex information

A quadratic with vertex $(h,k)$ can be written as $a(x-h)^2+k$ . Here, the vertex is $(3,21)$ .

Find the missing coefficient

Substitute one of the given points, such as $(0,12)$ , into your vertex-form equation to solve for $a$ .

Plug in $x=1$

After you know $a$ , substitute $x=1$ to compute $h(1)$ .

Desmos Guide

Enter the vertex-form model

In Desmos, type

h(x)=a(x-3)^2+21

Use a point to solve for $a$

Type the equation

12 = a(0-3)^2 + 21

and solve for $a$ (Desmos will show $a=-1$ ).

Evaluate at $x=1$

Type h(1) to display the value of $h(1)$ .

Step-by-step Explanation

Write the quadratic in vertex form

Because the highest point (vertex) is $(3,21)$ , write

h(x)=a(x-3)^2+21.

Use a point to find $a$

Substitute the point $(0,12)$ :

12=a(0-3)^2+21=9a+21.

So $9a=-9$ , and therefore $a=-1$ .

Evaluate $h(1)$

Now compute

h(1)=-1(1-3)^2+21=-4+21=17.

Therefore, $h(1)=17$ .

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