a rock is launched from a cannon. its height, h(x), can be represented by a quadratic function in terms of…

Explanation:

Step1: Set up the quadratic function

Let $h(x)=ax^{2}+bx + c$. When $x = 1$, $h(1)=a\times1^{2}+b\times1 + c=a + b + c=129$. When $x = 2$, $h(2)=a\times2^{2}+b\times2 + c=4a+2b + c=236$. Also, when $x = 0$ (at the moment of launch), $h(0)=c$ (the initial - height, assume the cannon is at ground - level, so $c = 0$). Then the system of equations becomes $\begin{cases}a + b=129\4a+2b=236\end{cases}$.

Step2: Solve the system of equations

From the first equation $b = 129 - a$. Substitute $b = 129 - a$ into the second equation: $4a+2(129 - a)=236$. Expand: $4a + 258-2a=236$. Combine like - terms: $2a=236 - 258=-22$, so $a=-11$. Then $b=129-(-11)=140$. So the height function is $h(x)=-11x^{2}+140x$.

Step3: Find the height at $x = 5$

Substitute $x = 5$ into $h(x)$: $h(5)=-11\times5^{2}+140\times5=-11\times25 + 700=-275+700 = 425$.

Answer:

425