For some reason, we're missing t in the equation. It should be:
d(t) = 96 + 80t - 16t²
A) To find when the ball reaches its highest point, derive with respect to t, and set that derivative equal to zero:
d(t) = 96 + 80t - 16t² Find the derivative, d'(t)
d'(t) = 80 - 32t Set d'(t) = 0
0 = 80 - 32t Add 32t to both sides
32t = 80 Divide both sides by 32
t = 2.5
Ball reaches max height at t=2.5 seconds
What is the max height? Plug t=2.5 into the original equation:
d(t) = 96 + 80t - 16t²
d(2.5) = 96 + 80(2.5) - 16(2.5²)
d(2.5) = 196
2)
a) A rational function that has a vertical asymptote at -5 and a hole at 7.
y = (x-7) ÷ (x-7)(x + 5)
In this function, there is a vertical assymptote at x=-5. As c approaches -5 from either direction, y approaches positive/negative infinite.
The hole is at 7. The function is undefined at 7, since it creates a zero in the denominator. Since (x-7) is also in the numerator, however, there is no asymptote.
b) A slant asymptote, or oblique/diagonal asymptote, occurs when the numerator contains a higher degree polynomial than the denomonator. The quotient polynomial gives the equation for the slant asymptote.
It's tough to think of an example of a polynomial function with a variable in both the numerator and denominator in which the denominator never equals zero. Here's a cheap answer: the denominator equals 1! A linear function does indeed have a slant asymptote.
here's our rather sly example:
y = 2x + 7
