Find the intercepts of the function?

Question

I can find the asymptotes no problem, but having trouble with the intercepts
G(x)= (x+2) / x^2+2x-3
G(x)= (x^2-3x-7) / (x+3)

Answer 1

x-intercepts are when the numerator = 0
x + 2 = 0 so x = -2 is an x-intercept (-2,0)

y-intercept is found by putting a zero in for x
G(0) = (0 + 2)/(0 + 0 -3) = -2/3 so y-int. at (0, -2/3)

second problem:
x-int: x^2 - 3x - 7 = 0 doesn't factor, so use the quadratic formula
x = (3 +/- sqrt(9 + 28))/2 = [3 +/- sqrt(37)]/2
approx with a calculator: x = -1.54 and x = 4.54
(-1.54, 0) and (4.54, 0)

y-int: G(0) = (0 - 0 - 7)/(0 + 3) = -7/2 so y-int at (0, -7/3)

Answer 2

The y-intercept is found by making x = 0 and solve for y.
The x-intercept is found by making y = 0 (or G(x) = 0) and solving for x.

1)

G(x) = (x + 2) / (x^2 + 2x - 3)

First, the y-intercept; make x = 0.

G(0) = (0 + 2) / (0^2 + 2*0 - 3)
G(0) = 2/(0 + 0 - 3)
G(0) = 2/(-3)
G(0) = -2/3

Therefore, the y-intercept is -2/3.

To find the x-intercepts, make G(x) = 0.

0 = (x + 2) / (x^2 + 2x - 3)

This is only true if the numerator is 0; that is, if
(x + 2) = 0, or
x = -2

Therefore, the x-intercept is -2.

Solve #2 on your own using the same concept.

Answer 3

the y intercepts is when x = 0
so G(0) = -2/3
G(0) = -7/3

and for the x intercept g(x) = 0
so x = -2

the second one you have to solve for the x^2 - 3x -7 = 0
and find the two real root