Question

Find the vertex, x intercept, and y intercept of the following equation: y=(x-3)^2+5

Answer 1

Answer:

The x-intercepts are:    ( 3 + i√5, 0) and  (3 - i√5, 0).

The y-intercept is (0,5).

The vertex is at (3, 5).

Step-by-step explanation:

The given equation y=(x-3)^2+5 has the general form y=(x-h)^2+k, whose vertex is at (h,k).

Thus, the vertex of y=(x-3)^2+5 is (3,5).

Find the x-intercepts by setting y=(x-3)^2+5 = 0 and solving for x.  To do this, solve (x-3)^2+5 = 0 for x.  Subtracting 5 from both sides, we get:

(x-3)^2 = -5.  We want to solve this for x.

Taking the square root of both sides, we get:

x-3 = √-1 * √5. Then x = 3 + i√5 and x = 3 - i√5.  The x-intercepts are:

( 3 + i√5, 0) and  (3 - i√5, 0).