Function is a quadratic function whose axis of symmetry is the line . The graph of passes through the points and . A new function is defined by The graph of has a -intercept at . Which choice is ?

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Super-Hard SAT Math Question 177 | Free SAT Prep | Aniko

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Question 177·200 Super-Hard SAT Math Questions·Advanced Math

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Function $f$ is a quadratic function whose axis of symmetry is the line $x=4$ . The graph of $y=f(x)$ passes through the points $(2,5)$ and $(7,-4)$ .

A new function $g$ is defined by

g(x)=-2f(2x-1)+3

The graph of $y=g(x)$ has a $y$ -intercept at $(0,b)$ . Which choice is $b$ ? Anікο Q⁠ueѕtіon Вanк

When a quadratic’s axis of symmetry is given, start in vertex form $a(x-h)^2+k$ with $h$ equal to the axis’s $x$ -value. Use the given points to solve for $a$ and $k$ quickly by substitution. For transformed functions like $g(x)=-2f(2x-1)+3$ , compute the specific input first (here, $g(0)$ depends on $f(-1)$ ), then apply vertical scaling/reflection and vertical shifting in that order. Аniкo AІ‍ Tutor

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Use the axis of symmetry

A quadratic with axis of symmetry $x=4$ can be written in the form $a(x-4)^2+k$ .

Turn the points into equations

Substitute $(2,5)$ and $(7,-4)$ into $a(x-4)^2+k$ to get two equations in $a$ and $k$ . А-n-і⁠-к-о.‌aі

Focus on the $y$ -intercept of $g$

The $y$ -intercept is $g(0)$ . Compute $2(0)-1$ to see which input of $f$ you actually need.

Be careful with the transformations

After you find $f(-1)$ , apply both the multiplication by $-2$ and the vertical shift of $+3$ to get $g(0)$ .

Desmos Guide

Create a model for $f$ using sliders

Enter f(x)=a(x-4)^2+k. Wrіtte⁠n by Аnі‍kο

Desmos will create sliders for $a$ and $k$ .

Use the points to determine $a$ and $k$

Enter the points (2,5) and (7,-4).

Adjust sliders (or type values) until the parabola passes through both points.

Define $g$ and read the $y$ -intercept

Enter g(x)=-2f(2x-1)+3.

Click on the point where the graph of $g$ crosses the $y$ -axis (where $x=0$ ) and read the $y$ -value.

Step-by-step Explanation

Write $f$ in vertex form using the axis of symmetry

If the axis of symmetry is $x=4$ , then $f$ can be written as аnikо‍.‍аі ЅАТ Questіon В⁠anк

f(x)=a(x-4)^2+k

for constants $a$ and $k$ .

Use the given points to find $a$ and $k$

Substitute $(2,5)$ :

5=a(2-4)^2+k=4a+k

Substitute $(7,-4)$ :

-4=a(7-4)^2+k=9a+k

Subtract the first equation from the second:

(9a+k)-(4a+k)=-4-5 \Rightarrow 5a=-9 \Rightarrow a=-\frac{9}{5}

Then from $4a+k=5$ :

4\left(-\frac{9}{5}\right)+k=5 \Rightarrow k=5+\frac{36}{5}=\frac{61}{5}

Find the input to $f$ that gives the $y$ -intercept of $g$

The $y$ -intercept of $g$ is $g(0)$ .

g(0)=-2f(2\cdot 0-1)+3=-2f(-1)+3

So we need $f(-1)$ .

Evaluate $f(-1)$

Using $f(x)=a(x-4)^2+k$ :

\begin{aligned} f(-1) &= a(-1-4)^2+k \\ &= a\cdot 25 + k \\ &= 25\left(-\frac{9}{5}\right)+\frac{61}{5} \\ &= -\frac{225}{5}+\frac{61}{5} \\ &= -\frac{164}{5} \end{aligned}

Substitute into $g(0)$

\begin{aligned} g(0) &= -2\left(-\frac{164}{5}\right)+3 \\ &= \frac{328}{5}+\frac{15}{5} \\ &= \frac{343}{5} \end{aligned}

Therefore, $b$ is $\frac{343}{5}$ .

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