The asymptotes of the function g(x)=(x+2)/(x^2+2x-3) are...
a. vertical: x=-2, x=1; horizontal: y=0
b. vertical: x=-3, x=1; horizontal: y=0
c. vertical: x=-2, x=-3; horizontal: y=0
d. vertical: x=-3, x=1; horizontal: none
vertical asymptotes exist when the denominator is zero, so
x^2 + 2x - 3 = 0
(x+3)(x-1) = 0
x = -3, x = 1
as x ---> infinity , g(x) = large/even larger ---> 0
y = 0
so it looks like b)
THANKS!
To find the vertical asymptotes of the function g(x) = (x+2)/(x^2+2x-3), we need to determine the values of x that cause the denominator to equal zero. Vertical asymptotes occur when the denominator of a rational function equals zero.
The denominator of g(x) is x^2 + 2x - 3. To find the values of x that make the denominator zero, we can solve the equation x^2 + 2x - 3 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring this particular equation may not be straightforward, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For the equation x^2 + 2x - 3 = 0, a = 1, b = 2, and c = -3.
Plugging these values into the quadratic formula, we have:
x = (-2 ± √(2^2 - 4(1)(-3))) / (2(1))
x = (-2 ± √(4 + 12)) / 2
x = (-2 ± √16) / 2
x = (-2 ± 4) / 2
Simplifying further, we have:
x = -1 ± 2
This gives us two possible values for x: x = -3 and x = 1.
Therefore, the vertical asymptotes of the function g(x) = (x+2)/(x^2+2x-3) are x = -3 and x = 1.
For the horizontal asymptote, we need to determine what happens to the function as x approaches positive or negative infinity. To do this, we can compare the degrees of the numerator and denominator. The degree of the numerator is 1, and the degree of the denominator is 2.
Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
Therefore, the correct answer is option a. Vertical asymptotes: x = -2, x = 1; horizontal asymptote: y = 0.