Let the function p be defined as p(x) = ((x - c)^2 + 160)/(2c) , where c is a constant. If p(c) = 10 , what is the value of p(12) ?

SAT Question 1a722d7d: Answer, Explanation & Desmos Solution

Let the function $p$ be defined as $p (x) = \frac{( x - c ) ^{2} + 160}{2 c}$ , where $c$ is a constant. If $p (c) = 10$ , what is the value of $p (12)$ ?

Difficulty:Hard

$10.00$

$10.25$

$10.75$

$11.00$

Explanation

Correct answer: D.

Choice D is correct. The value of $p (12)$ depends on the value of the constant $c$ , so the value of $c$ must first be determined. It is given that $p (c) = 10$ . Based on the definition of $p$ , it follows that:

$p (c) = \frac{( c - c ) ^{2} + 160}{2 c} = 10$

$\frac{160}{2 c} = 10$

$2 c = 16$

$c = 8$

This means that $p (x) = \frac{( x - 8 ) ^{2} + 160}{16}$ for all values of $x$ . Therefore:

$p (12) = \frac{( 12 - 8 ) ^{2} + 160}{16}$

$= \frac{16 + 160}{16}$

$= 11$

Desmos solution:

Enter in Desmos:

$p (x) = \frac{( x - c ) ^{2} + 160}{2 c}$
$p (c) \sim 10 {c > 0}$
$p (12)$

Reviewed and rewritten by CookSAT

Why the other choices are wrong

Choice A

Choice A is incorrect. It is the value of $p (8)$ , not $p (12)$ .

Choice B

Choice B is incorrect. If this value were correct, then $x = 12$ and the selected value of $p (12)$ could be substituted into the equation to solve for $c$ . However, the value of $c$ that results from choice B results in $p (c) < 10$ .

Choice C

Choice C is incorrect. If this value were correct, then $x = 12$ and the selected value of $p (12)$ could be substituted into the equation to solve for $c$ . However, the value of $c$ that results from choice C results in $p (c) < 10$ .

Student discussion

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SAT Question 1a722d7d: Nonlinear Functions answer and explanation with Desmos solution

Explanation

Enter in Desmos:

Why the other choices are wrong

Student discussion