Function is a quadratic function. The graph of in the -plane has a vertex at and passes through the point . The graph of has a -intercept at . A new function is defined by . The graph of has a -intercept at . Which choice is the positive difference between and ?

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Super-Hard SAT Math Question 129 | Free SAT Prep | Aniko

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Question 129·200 Super-Hard SAT Math Questions·Advanced Math

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Function $f$ is a quadratic function. The graph of $y=f(x)$ in the $xy$ -plane has a vertex at $(-2,7)$ and passes through the point $(1,-5)$ . The graph of $y=f(x)$ has a $y$ -intercept at $(0,a)$ .

A new function $p$ is defined by

$p(x)=\frac{1}{2}f(3x+6)-4$ .

The graph of $y=p(x)$ has a $y$ -intercept at $(0,b)$ . Which choice is the positive difference between $a$ and $b$ ?

When a quadratic’s vertex is given, use vertex form immediately: $f(x)=c(x-h)^2+k$ . One additional point determines $c$ quickly. For transformed functions like $p(x)=\tfrac12 f(3x+6)-4$ , remember that a $y$ -intercept means plugging in $x=0$ , which changes the input to $f$ (here it becomes $6$ ). Finish by carefully combining fractions with a common denominator and taking the absolute value for a positive difference.

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Start with vertex form

Because the vertex is given, write $f(x)$ as $c(x+2)^2+7$ and use the point $(1,-5)$ to find $c$ .

Be careful with the input to $f$ in $p(0)$

To find the $y$ -intercept of $p$ , compute $p(0)$ . That requires evaluating $f(3\cdot 0+6)$ .

Use common denominators for the final difference

When you compute $|a-b|$ , rewrite both numbers with the same denominator before adding or subtracting.

Desmos Guide

Define the quadratic from the vertex and point

Enter c = (-5-7)/9 to compute the coefficient from $-5 = c(1+2)^2+7$ .

Then enter f(x)=c(x+2)^2+7.

Compute a and b from the y-intercepts

Enter a=f(0).

Enter b=0.5*f(6)-4 (since $p(0)=\tfrac12 f(6)-4$ ).

Compute the positive difference

Enter d=abs(a-b).

Match the displayed value of d to one of the answer choices.

Step-by-step Explanation

Use vertex form

A quadratic with vertex $(-2,7)$ can be written as

$f(x)=c(x+2)^2+7$

for some constant $c$ .

Find the coefficient using the point

Since the graph passes through $(1,-5)$ ,

\begin{aligned} -5 &= c(1+2)^2+7 \\ -5 &= 9c+7 \\ -12 &= 9c \\ c &= -\frac{4}{3} \end{aligned}

So $f(x)=-\frac{4}{3}(x+2)^2+7$ .

Compute the y-intercept of $f$

The $y$ -intercept occurs at $x=0$ :

\begin{aligned} a=f(0) &= -\frac{4}{3}(0+2)^2+7 \\ &= -\frac{4}{3}\cdot 4+7 \\ &= -\frac{16}{3}+\frac{21}{3} \\ &= \frac{5}{3} \end{aligned}

Compute the y-intercept of $p$

The $y$ -intercept of $p$ occurs at $x=0$ :

\begin{aligned} b=p(0) &= \frac{1}{2}f(3\cdot 0+6)-4 \\ &= \frac{1}{2}f(6)-4 \end{aligned}

Now compute $f(6)$ :

\begin{aligned} f(6) &= -\frac{4}{3}(6+2)^2+7 \\ &= -\frac{4}{3}\cdot 64+7 \\ &= -\frac{256}{3}+\frac{21}{3} \\ &= -\frac{235}{3} \end{aligned}

\begin{aligned} b &= \frac{1}{2}\left(-\frac{235}{3}\right)-4 \\ &= -\frac{235}{6}-\frac{24}{6} \\ &= -\frac{259}{6} \end{aligned}

Take the positive difference

\begin{aligned} |a-b| &= \left|\frac{5}{3}-\left(-\frac{259}{6}\right)\right| \\ &= \left|\frac{10}{6}+\frac{259}{6}\right| \\ &= \frac{269}{6} \end{aligned}

So the correct choice is $269/6$ .

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