<em><u>Question:</u></em>
Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)
<em><u>Answer:</u></em>
The vertical asymptotes are x = 2 or x = -1
<em><u>Solution:</u></em>
Given that,
For vertical asymptotes, set the denominator to zero and then solve the quadratic equation
Thus the vertical asymptotes are x = - 1 and x = 2
Answer:
9/2
Step-by-step explanation:
this is a simple integral function
given limits of interval [a,b] of a continuous function f(t), you can find the area under the curve by using:
using the fundamental theorem of calculus that states the integral of f(x) in the interval [a,b] is = g(a)-g(b), where g(x) is the antiderivative of f(x)
our g(x) =
g(3)-g(0) = g(3) = 27/2 - 27/3 = 27/2-9 = 9/2