| | | |---:|---:| | | | | | | | | | Three points on the quadratic function are given in the table. If , which choice is the -coordinate of the -intercept of the function ?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 82 | Free SAT Prep | Aniko

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Question 82·200 Super-Hard SAT Math Questions·Advanced Math

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$x$	$f(x)$
$-4$	$-3$
$2$	$0$
$6$	$7$

Three points on the quadratic function $f(x)$ are given in the table. If $g(x)=f(x-3)$ , which choice is the $y$ -coordinate of the $y$ -intercept of the function $g(x)$ ?

For transformation questions, immediately rewrite what the question is really asking: the $y$ -intercept is $g(0)$ , and with $g(x)=f(x-3)$ that becomes $f(-3)$ . Then use the three given points to determine $f(x)$ (often via $ax^2+bx+c$ and elimination), and finally substitute the required $x$ -value.

Hints

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Use the definition of y-intercept

The $y$ -intercept of $g$ is $g(0)$ . Compute $g(0)$ from $g(x)=f(x-3)$ to see which input of $f$ you need.

Model the quadratic in standard form

Since you have 3 points, write $f(x)=ax^2+bx+c$ and plug each point into the equation to get three equations.

Eliminate a variable

If you subtract one equation from another, the $c$ terms cancel. This makes it easier to solve for $a$ and $b$ first.

Finish solving for $c$

After finding $a$ and $b$ , substitute back into any of the three original equations to get $c$ .

Evaluate at the needed x-value

Once you know $f(x)$ , compute $f(-3)$ (because $g(0)=f(-3)$ ).

Desmos Guide

Enter the points as lists

In Desmos, enter:

x1={-4,2,6}
y1={-3,0,7}

Fit an exact quadratic model

Type:

y1 ~ a x1^2 + b x1 + c

With 3 points, Desmos will determine exact values for $a$ , $b$ , and $c$ .

Evaluate the needed input

Because $g(0)=f(-3)$ , evaluate the model at $x=-3$ by typing:

a(-3)^2 + b(-3) + c

Interpret the result

The value returned is $f(-3)$ , which equals $g(0)$ and is the $y$ -coordinate of the $y$ -intercept of $g(x)$ .

Step-by-step Explanation

Relate the y-intercept of $g$ to a value of $f$

The $y$ -intercept of $g$ occurs at $x=0$ , so compute

g(0)=f(0-3)=f(-3).

So the problem reduces to finding $f(-3)$ .

Write a general quadratic and plug in the points

Let

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

Use the table:

From $(-4,-3)$ :

16a-4b+c=-3

From $(2,0)$ :

4a+2b+c=0

From $(6,7)$ :

36a+6b+c=7

Eliminate $c$ to solve for $a$ and $b$

Subtract the $(2,0)$ equation from the $(6,7)$ equation:

(36a+6b+c)-(4a+2b+c)=7-0 \Rightarrow 32a+4b=7.

Subtract the $(2,0)$ equation from the $(-4,-3)$ equation:

(16a-4b+c)-(4a+2b+c)=-3-0 \Rightarrow 12a-6b=-3.

Simplify:

\begin{aligned} 8a+b&=\frac{7}{4} \\ 2a-b&=-\frac{1}{2} \end{aligned}

Find $a$ , $b$ , and $c$

Add the two simplified equations:

(8a+b)+(2a-b)=\frac{7}{4}-\frac{1}{2} \Rightarrow 10a=\frac{5}{4} \Rightarrow a=\frac{1}{8}.

Then

b=\frac{7}{4}-8a=\frac{7}{4}-1=\frac{3}{4}.

Use $4a+2b+c=0$ to find $c$ :

4\left(\frac{1}{8}\right)+2\left(\frac{3}{4}\right)+c=0 \Rightarrow \frac{1}{2}+\frac{3}{2}+c=0 \Rightarrow c=-2.

Evaluate $f(-3)$ to get the y-intercept of $g$

Now compute

f(-3)=a(-3)^2+b(-3)+c.

Substitute $a=\frac{1}{8}$ , $b=\frac{3}{4}$ , $c=-2$ :

\begin{aligned} f(-3)&=\frac{1}{8}\cdot 9+\frac{3}{4}(-3)-2 \\ &=\frac{9}{8}-\frac{9}{4}-2 \\ &=\frac{9}{8}-\frac{18}{8}-\frac{16}{8} \\ &=-\frac{25}{8}. \end{aligned}

Therefore, the $y$ -coordinate of the $y$ -intercept of $g(x)$ is $-\frac{25}{8}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

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Strategy

Hints

Desmos Guide

Step-by-step Explanation