Question 802·Hard·Math

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The polynomial function

p(x)=k\,(x+3)^{2}(x-5)

is defined above, where $k$ is a nonzero constant. If $p(1)=192$ , what is the value of $p(0)$ ?

Your answer

A parameter is a fixed but unknown number in a function's rule. A given output at one input can determine it, even when the question asks for an output at another input. Find the constant from the known output, then evaluate at the requested input. Use a regression in Desmos to find the constant, but don't mistake it for the answer.

Hints

Hint 1
An input is the number inside the function's parentheses. To use $p(1)=192$ , replace every $x$ in the rule with $1$ . Which part of the resulting equation is still unknown?
Hint 2
A regression lets Desmos find a constant that makes an equation true. Type the equation for $p(1)$ with $\sim$ in place of $=$ . Since $1-5$ is negative, what sign must $k$ have to give a positive output?
Hint 3
Once Desmos finds $k$ , define the given function and type $p(0)$ . Evaluating $p(0)$ means replacing $x$ with $0$ , not replacing $k$ with $0$ .

Step-by-step

Find the constant, then evaluate
Step 1
Turn the known output into an equation
$p(1)=192$ means the output is $192$ when the input is $1$ . Replace every $x$ in the given rule with $1$ :
$192=k(1+3)^2(1-5).$
Only $k$ , the fixed constant, is unknown.
Step 2
Find the constant in Desmos
Type $192\sim k(1+3)^2(1-5)$ . The regression symbol $\sim$ tells Desmos to find the value of $k$ that fits the known output. Under PARAMETERS, Desmos shows $k=-3$ . That sign makes sense: $1-5$ is negative, and the given output is positive.
Step 3
Define the function with the found constant
Type $p(x)=k(x+3)^2(x-5)$ . Desmos uses the $k$ it found, so this is the same function as the one in the question, now ready for a new input.
Step 4
Evaluate the requested output
Type $p(0)$ , which tells Desmos to put $0$ in for $x$ . Desmos shows $135$ . That's the value of $p(0)$ , not the value of $k$ . Grid in 135.

Question 802·Hard·Math

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