The function is defined by , where is a positive constant. The graph of has a minimum point that lies on the line . Which choice is the value of ?

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SAT Nonlinear Functions Question 59 | Free SAT Prep | Aniko

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Question 59·Hard·Nonlinear Functions

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The function $f$ is defined by $f(x)=(x+3)(x-c)$ , where $c$ is a positive constant. The graph of $y=f(x)$ has a minimum point that lies on the line $y=2x-25$ .

Which choice is the value of $c$ ?

When a quadratic is given in factored form, use structure instead of expanding: the zeros are visible, so the vertex x-coordinate is the average of the zeros. Then evaluate the function at that x-value using the factors (often it simplifies nicely). Finally, translate any geometric condition like “the vertex lies on a line” into an equation by substituting the vertex coordinates.

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Use the zeros

From $f(x)=(x+3)(x-c)$ , identify the two x-intercepts of the parabola.

Find the vertex from symmetry

The x-coordinate of the vertex is halfway between the two zeros.

Turn the “vertex lies on a line” fact into an equation

Compute the vertex as $(h,f(h))$ in terms of $c$ , then substitute into $y=2x-25$ to get one equation in $c$ .

Desmos Guide

Create an equation in one variable

Define a function in terms of a single variable (use $x$ to represent $c$ ):

g(x)= -((x+3)^2)/4 - (x-28)

This comes from setting the vertex y-value $-\frac{(c+3)^2}{4}$ equal to the line value $c-28$ .

Graph the function

Graph y=g(x).

Find where the equation is satisfied

Find the x-intercepts of the graph (where $g(x)=0$ ).

Choose the valid intercept

There will be two x-intercepts. Since $c$ must be positive, select the positive x-intercept.

Step-by-step Explanation

Write the vertex x-coordinate using symmetry

The zeros of $f(x)=(x+3)(x-c)$ are $x=-3$ and $x=c$ .

For a parabola, the vertex x-coordinate is the average of the zeros:

h=\frac{-3+c}{2}=\frac{c-3}{2}

Find the vertex y-coordinate in terms of $c$

Evaluate $f(h)$ at $h=\frac{c-3}{2}$ .

First compute each factor:

\begin{aligned} h+3 &= \frac{c-3}{2}+3=\frac{c+3}{2} \\ h-c &= \frac{c-3}{2}-c=\frac{-c-3}{2}=-\frac{c+3}{2} \end{aligned}

So the vertex y-coordinate is

k=f(h)=(h+3)(h-c)=\frac{c+3}{2}\left(-\frac{c+3}{2}\right)=-\frac{(c+3)^2}{4}

Use the fact that the vertex lies on the given line

Because the minimum point (vertex) lies on $y=2x-25$ , substitute $x=h$ and $y=k$ :

-\frac{(c+3)^2}{4}=2\left(\frac{c-3}{2}\right)-25

Simplify the right side:

2\left(\frac{c-3}{2}\right)-25=(c-3)-25=c-28

-\frac{(c+3)^2}{4}=c-28

Solve for $c$ and choose the positive value

Multiply both sides by $4$ :

-(c+3)^2=4(c-28)

Expand and rearrange:

\begin{aligned} -(c^2+6c+9) &= 4c-112 \\ -c^2-6c-9 &= 4c-112 \\ 0 &= c^2+10c-103 \end{aligned}

Use the quadratic formula:

\begin{aligned} c &= \frac{-10\pm\sqrt{10^2-4(1)(-103)}}{2} \\ &= \frac{-10\pm\sqrt{512}}{2} \\ &= \frac{-10\pm 16\sqrt{2}}{2} \\ &= -5\pm 8\sqrt{2} \end{aligned}

Since $c$ is positive, the correct choice is $-5+8\sqrt{2}$ .

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Step-by-step Explanation

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