Question

1. pumpkin - launching contest the path of a pumpkin launched from a compressed air cannon is modeled by (f(x)=-\frac{3}{10,000}x^{2}+\frac{3}{2}x + 24) where (f(x)) is the height (in feet) and (x) is the horizontal distance (in feet) from where the pumpkin was launched.

a. how high is the pumpkin when it is launched?
b. what is the maximum height of the pumpkin?

Explanation:

Step1: Find height when launched

When the pumpkin is launched, the horizontal distance $x = 0$. Substitute $x = 0$ into the function $f(x)=-\frac{3}{10000}x^{2}+\frac{3}{2}x + 24$.
$f(0)=-\frac{3}{10000}(0)^{2}+\frac{3}{2}(0)+24$

Step2: Calculate the value

$f(0)=24$

Step3: Find the x - coordinate of the vertex for maximum height

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. Here, $a =-\frac{3}{10000}$ and $b=\frac{3}{2}$.
$x=-\frac{\frac{3}{2}}{2\times(-\frac{3}{10000})}=\frac{\frac{3}{2}}{\frac{6}{10000}}=\frac{3}{2}\times\frac{10000}{6}= 2500$

Step4: Find the maximum height

Substitute $x = 2500$ into the function $f(x)=-\frac{3}{10000}x^{2}+\frac{3}{2}x + 24$.
$f(2500)=-\frac{3}{10000}(2500)^{2}+\frac{3}{2}(2500)+24$
$=-\frac{3}{10000}\times6250000+\frac{7500}{2}+24$
$=- 1875 + 3750+24$
$=1899$

Answer:

a. 24 feet
b. 1899 feet