The function is defined by where is a quadratic polynomial. The table shows three values of and the corresponding values of . | | | |----|----| | | | | | | | | | What are the coordinates of the vertex of the graph of in the -plane?

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SAT Nonlinear Functions Question 219 | Free SAT Prep | Aniko

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Question 219·Hard·Nonlinear Functions

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The function $q$ is defined by

q(x)=\frac{p(x)-5}{x+1},

where $p$ is a quadratic polynomial. The table shows three values of $x$ and the corresponding values of $q(x)$ .

$x$	$q(x)$
$2$	$3$
$4$	$7$
$-4$	$-1$

What are the coordinates of the vertex of the graph of $y=p(x)$ in the $xy$ -plane?

On this type of SAT problem, first rewrite the given relation so the polynomial is isolated, here turning $q(x) = \dfrac{p(x)-5}{x+1}$ into $p(x) = (x+1)q(x)+5$ and quickly computing $p(x)$ at the specified $x$ -values. Then model $p(x)$ as $ax^2 + bx + c$ , use the three resulting points to form and solve a small linear system for $a$ , $b$ , and $c$ , and apply the vertex formula $x = -\tfrac{b}{2a}$ plus one substitution to get the vertex coordinates. This method is systematic, avoids guessing, and keeps the algebra manageable under time pressure.

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Use the definition of q(x) to get p(x) values

Start by isolating $p(x)$ in $q(x) = \dfrac{p(x) - 5}{x + 1}$ . How can you rewrite this so $p(x)$ is alone on one side?

Turn q-values into p-values

Once you have $p(x)$ in terms of $x$ and $q(x)$ , plug in each $x$ and its table value of $q(x)$ to compute $p(2)$ , $p(4)$ , and $p(-4)$ .

Model p(x) as a quadratic and form equations

Assume $p(x) = ax^2 + bx + c$ and use the three $(x, p(x))$ pairs you found to write three equations in $a$ , $b$ , and $c$ .

Use the vertex formula

After you find $a$ , $b$ , and $c$ , recall that the vertex of $y = ax^2 + bx + c$ has $x$ -coordinate $-\dfrac{b}{2a}$ . Then substitute this $x$ back into $p(x)$ to get the $y$ -coordinate.

Desmos Guide

Convert q-values to p-values using Desmos

In separate expression lines, type:

(2+1)*3+5
(4+1)*7+5
(-4+1)*(-1)+5 Desmos will output three numbers; these are $p(2)$ , $p(4)$ , and $p(-4)$ and give you three points on the parabola $y=p(x)$ .

Enter the three points for p(x)

Create a table and enter the $x$ -values in the first column (for example, 2, 4, -4) and the corresponding $p(x)$ values you just found in the second column. These points lie on the graph of $y = p(x)$ .

Fit a quadratic to the points

In a new expression line, type y1 ~ a x1^2 + b x1 + c to perform a quadratic regression. Desmos will display values for a, b, and c; this gives you the equation $y = ax^2 + bx + c$ for $p(x)$ .

Find the vertex from the quadratic

Graph the function y = a x^2 + b x + c using the values of a, b, and c. Either:

Click on the lowest point of the parabola to see its coordinates (the vertex), or
Add expressions x_v = -b/(2a) and y_v = a x_v^2 + b x_v + c and read off $(x_v, y_v)$ from the graph or the expression list. The coordinate pair you see is the vertex of $y=p(x)$ .

Step-by-step Explanation

Relate p(x) and q(x) and find three points on p

From the definition

q(x)=\frac{p(x)-5}{x+1},

solve for $p(x)$ :

p(x) = (x+1)q(x) + 5.

Use the table values to compute $p(x)$ :

For $x=2$ : $p(2) = (2+1)\cdot3 + 5 = 3\cdot3 + 5 = 14$ .
For $x=4$ : $p(4) = (4+1)\cdot7 + 5 = 5\cdot7 + 5 = 40$ .
For $x=-4$ : $p(-4) = (-4+1)\cdot(-1) + 5 = (-3)(-1) + 5 = 8$ .

So the parabola $y=p(x)$ passes through the points $(2,14)$ , $(4,40)$ , and $(-4,8)$ .

Write a general quadratic for p(x) and set up equations

Let

p(x) = ax^2 + bx + c.

Use the three points you just found to create equations:

From $(2,14)$ : $4a + 2b + c = 14$ .
From $(4,40)$ : $16a + 4b + c = 40$ .
From $(-4,8)$ : $16a - 4b + c = 8$ .

Now you have a system of three linear equations in $a$ , $b$ , and $c$ .

Solve the system for a, b, and c

Subtract the first equation from the second:

\begin{aligned} (16a + 4b + c) - (4a + 2b + c) &= 40 - 14 \\ 12a + 2b &= 26 \quad\Rightarrow\quad 6a + b = 13 \end{aligned}

Subtract the third equation from the second:

\begin{aligned} (16a + 4b + c) - (16a - 4b + c) &= 40 - 8 \\ 8b &= 32 \quad\Rightarrow\quad b = 4 \end{aligned}

Plug $b = 4$ into $6a + b = 13$ :

6a + 4 = 13 \Rightarrow 6a = 9 \Rightarrow a = \frac{3}{2}.

Use $4a + 2b + c = 14$ to find $c$ :

\begin{aligned} 4\cdot\frac{3}{2} + 2\cdot4 + c &= 14 \\ 6 + 8 + c &= 14 \quad\Rightarrow\quad c = 0 \end{aligned}

p(x) = \frac{3}{2}x^2 + 4x.

Find the vertex of p(x)

For a quadratic $p(x) = ax^2 + bx + c$ , the $x$ -coordinate of the vertex is

x_v = -\frac{b}{2a}.

Here $a = \dfrac{3}{2}$ and $b = 4$ , so

x_v = -\frac{4}{2\cdot(3/2)} = -\frac{4}{3}.

Now find $y_v = p(x_v)$ :

p\left(-\frac{4}{3}\right) = \frac{3}{2}\left(-\frac{4}{3}\right)^2 + 4\left(-\frac{4}{3}\right) = \frac{3}{2}\cdot\frac{16}{9} - \frac{16}{3} = \frac{8}{3} - \frac{16}{3} = -\frac{8}{3}.

So the vertex of $y=p(x)$ is $\left(-\dfrac{4}{3}, -\dfrac{8}{3}\right)$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation