Question 123·200 Super-Hard SAT Math Questions·Advanced Math

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$g(x)$ is a quadratic function. In the $xy$ -plane, the graphs of $y=g(x)$ and $y=2x+1$ intersect at the points whose $x$ -coordinates are $-1$ and $5$ .

If $g(2)=7$ , which choice is the value of $g(8)$ ?

If a quadratic meets a line at two known inputs, those inputs are zeros of the quadratic minus the line, not necessarily zeros of the quadratic itself. Factor that difference, then use the extra function value to find its scale with a Desmos regression. Add the line back before evaluating the requested input; leaving it out answers a different question.

Hints

Hint 1
At an intersection, both graphs have the same output for the same input. So subtracting the line’s output from the quadratic’s output gives zero at each intersection. What expression should you subtract from $g(x)$ ?
Hint 2
A zero is an input that makes an expression equal $0$ . Zeros at $-1$ and $5$ give factors $x+1$ and $x-5$ , but these belong to the difference between the functions, not to $g$ alone.
Hint 3
After you add the line back, $g(2)=7$ gives an equation with one unknown scale, $a$ . A Desmos regression can find $a$ ; then use its stored value to evaluate $g$ at the requested input.

Step-by-step

Factor the difference between the graphs
Step 1
Find what equals zero at the intersections
At an intersection, the two graphs have the same output. So $d(x)=g(x)-(2x+1)$ equals $0$ at both $x=-1$ and $x=5$ . The intersections give zeros of the difference, not zeros of $g$ .
Step 2
Factor the difference
The zeros $-1$ and $5$ give factors $x+1$ and $x-5$ : each factor becomes $0$ at its matching input. A quadratic with those zeros can have different scales, so write its unknown scale as $a$ : $d(x)=a(x+1)(x-5)$ .
Step 3
Recover the function you need
Since $d(x)$ was $g(x)$ minus the line, add the line back to get $g(x)$ : $g(x)=a(x+1)(x-5)+2x+1$ . Stopping at $d(x)$ would give the gap between the graphs, not the function asked for.
Step 4
Use the known output to make an equation
$g(2)=7$ means putting $2$ in for $x$ must give $7$ . Substitute it into the equation for $g$ : $7=a(2+1)(2-5)+2(2)+1$ . Neither factor is $0$ at $2$ , so this equation can determine $a$ .
Step 5
Find the missing scale in Desmos
Type $7\sim a(2+1)(2-5)+2(2)+1$ . The tilde tells Desmos to fit $a$ ; under PARAMETERS, it reports $a\approx-0.222222$ . Keep the negative sign: $2-5$ is negative.
Step 6
Evaluate the requested output
Keep that line and type $a(8+1)(8-5)+2(8)+1$ . Desmos uses its stored $a$ and prints $11$ . So $g(8)=11$ , not the line’s output alone. Choice D.

Question 123·200 Super-Hard SAT Math Questions·Advanced Math

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