Question 318·Medium·Math

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The function $h$ is defined by

h(x)=x^2-8x+12.

If $h(8)=h(t)$ and $t\neq 8$ , what is the value of $t$ ?

For a quadratic function, two different inputs can share one output. An equality like $h(a)=h(t)$ cues you to evaluate the known input in Desmos, then graph the function and that output as a horizontal line. Read the intersections as inputs, then apply any exclusion. The shared output and the parabola’s turning point are tempting answers, but neither is automatically the other input.

Hints

Hint 1
In $h(8)$ , the $8$ is the input. Put it into the function in Desmos to find the shared output. What output must the unknown input $t$ produce?
Hint 2
A horizontal line at the shared output crosses the parabola, the graph of a quadratic function, at every input that produces it. Graph both and look for their intersections.
Hint 3
Each intersection’s horizontal coordinate is an input that works. Use $t\neq 8$ to reject the input the question rules out, even though it gives the right output.

Step-by-step

Approach 1: Evaluate, then find the matching input
Step 1
Find the output at 8
Type $h(x)=x^2-8x+12$ , then $h(8)$ . The notation $h(8)$ means put $8$ in for $x$ . Desmos prints $12$ , so the output at $8$ is $12$ .
Step 2
Set the other output equal to 12
The given equality says the two inputs have the same output. Since $h(8)=12$ , the unknown input must satisfy $h(t)=12$ . Here $t$ is an input; $12$ is the output it must produce.
Step 3
Read both inputs and apply the restriction
Type $y=h(x)$ and $y=12$ . Click both intersections, where the graphs have the same height: $(0,12)$ and $(8,12)$ . Their horizontal coordinates give $t=0$ or $t=8$ . Because $t\neq 8$ , the other input is $t=0$ . Choice B.

Approach 2: Use the parabola’s symmetry
Step 1
Rewrite the function to show its center
No Desmos needed. Symmetry gives the other input exactly. Half of $-8$ is $-4$ , and its square is $16$ . Add and subtract $16$ to make a perfect square without changing the function: $h(x)=x^2-8x+16-16+12$ . Group the square: $h(x)=(x-4)^2-16+12$ . Combine the constants: $h(x)=(x-4)^2-4$ . The axis of symmetry, the line through the parabola’s vertex, is $x=4$ .
Step 2
Reflect the known input across the axis
Inputs equally far from $4$ have the same output because their distances are squared in $(x-4)^2$ . The input $8$ is $4$ units to the right of the axis, so its matching input is $4$ units to the left: $t=4-4=0$ . This input meets $t\neq 8$ , so $t=0$ . Choice B.

Question 318·Medium·Math

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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