Question

PLEASE Find the x and y-intercept(s) of = 2( + 1)^2 + 3.

Answer 1

To find the x-intercepts, we need to set y = 0 and solve for x:

2(x + 1)^2 + 3 = 0

2(x + 1)^2 = -3

(x + 1)^2 = -3/2

This equation has no real solutions because the square of any real number is always positive or zero, and -3/2 is negative.

To find the y-intercept, we need to set x = 0:

2(0 + 1)^2 + 3 = 5

Therefore, the y-intercept is (0, 5).