something like
y = (3x^2 + k)/((x+2)(x-2))
forcing (1,-1) to be on it
-1 = (3+k)/-3
3 = 3+k
k = 0
sketch y = (3x^2/(x^2 - 4)
using: www.desmos.com/calculator
Note:
f(x) = f(-x), so it is even
it has an asymptote at x = -2
(It did not say anything about having more than one vertical asymptote)
as x ---> large, f(x) ----> 3
it passes through (1, -1)
f(x) = quadratic/quadratic
all your conditions are met
Determine a function, f(x)=quadratic/quadratic, that satisfies the following.
-Even
-One vertical asymptote x=-2
-point (1,-1)
-Horizontal asymptote y=3
2 answers
Hmmm. The above function also has a vertical asymptote at x = 2, and a zero at x=0. I guess that fills the requirements, but not if you want to satisfy only those specifications
-One vertical asymptote x=-2
y = k/(x+2)
-Horizontal asymptote y=3
y = 3/(x+2)
-Even, quadratic/quadratic
y = 3(x^2+k)/(x+2)^2
- (1,-1)
3k/9 = -1
k = -3
y = 3(x^2-9)/(x+2)^2
But now we have introduced two zeroes, and the specifications do not say we should have any. But it turns out that we cannot have a horizontal asymptote at y=3 and pass through (1,-1) and not have it cross the x-axis somewhere (by the Intermediate Value Theorem). So I guess we can stop here.
-One vertical asymptote x=-2
y = k/(x+2)
-Horizontal asymptote y=3
y = 3/(x+2)
-Even, quadratic/quadratic
y = 3(x^2+k)/(x+2)^2
- (1,-1)
3k/9 = -1
k = -3
y = 3(x^2-9)/(x+2)^2
But now we have introduced two zeroes, and the specifications do not say we should have any. But it turns out that we cannot have a horizontal asymptote at y=3 and pass through (1,-1) and not have it cross the x-axis somewhere (by the Intermediate Value Theorem). So I guess we can stop here.
1 answer