The function is quadratic. If and , which choice is the value of ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 87 | Free SAT Prep | Aniko

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Question 87·200 Super-Hard SAT Math Questions·Advanced Math

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The function $f$ is quadratic. If $f(0)=1$ and $f(1)=f(5)=10$ , which choice is the value of $f(10)$ ?

When a quadratic gives equal outputs at two different inputs, use symmetry first: the axis of symmetry is the midpoint of those inputs. Then switch to vertex form $a(x-h)^2+k$ , which makes plugging in points fast and reduces messy expansion. After you solve for the parameters with two given values, substitute the requested input to get the final value.

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Look for symmetry

If a quadratic has the same output at two different $x$ -values, the axis of symmetry is halfway between those $x$ -values.

Use vertex form

After you find the axis of symmetry $x=h$ , write $f(x)=a(x-h)^2+k$ .

Plug in two given points

Use two of the given function values to create two equations in $a$ and $k$ , then solve and evaluate $f(10)$ .

Desmos Guide

Set up lines for the parameters

Let the horizontal axis represent $a$ and the vertical axis represent $k$ . Enter these two equations from substituting the given points into $f(x)=a(x-3)^2+k$ :

$k=1-9a$

$k=10-4a$

Find the intersection

Click the intersection point of the two lines. Its coordinates are $(a,k)$ for the quadratic.

Compute $f(10)$ from the parameters

In a new line, enter

$a(10-3)^2+k$

Then replace $a$ and $k$ with the intersection values (or read the computed value directly if you defined $a$ and $k$ as the intersection coordinates). The result is the correct answer choice.

Step-by-step Explanation

Use symmetry to locate the axis

For a quadratic function, if two $x$ -values have the same function value, they are equally spaced from the axis of symmetry.

Since $f(1)=f(5)$ , the axis of symmetry is at the midpoint:

\frac{1+5}{2}=3

So the vertex form can be written as $f(x)=a(x-3)^2+k$ .

Use the given points to solve for $a$ and $k$

Plug in $f(0)=1$ :

1=a(0-3)^2+k=9a+k

Plug in $f(1)=10$ :

10=a(1-3)^2+k=4a+k

Subtract the second equation from the first:

(9a+k)-(4a+k)=1-10

5a=-9 \Rightarrow a=-\frac{9}{5}

Then use $4a+k=10$ :

4\left(-\frac{9}{5}\right)+k=10 \Rightarrow -\frac{36}{5}+k=10 \Rightarrow k=\frac{86}{5}

Evaluate $f(10)$

Now compute:

\begin{aligned} f(10) &= a(10-3)^2+k \\ &= -\frac{9}{5}\cdot 49+\frac{86}{5} \\ &= \frac{-441+86}{5} \\ &= -\frac{355}{5} \\ &= -71 \end{aligned}

So the correct choice is -71.

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Step-by-step Explanation

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Step-by-step Explanation