Question

What is the vertex and y-intercept of y=-(x+3)^2+4

vertex: (−3, −4); y-intercept: 13


vertex: (3, −4); y-intercept: −9


vertex: (3, 4); y-intercept: −5


vertex: (−3, 4); y-intercept: −5

Answer 1

Vertex → (x, y) = (− 3, − 4)

x intercepts → (x, y) = (−5, 0) and (−1, 0)

y intercept → (x, y) = (0, 5)

Explanation:

Set y = 0 = ( x + 3)^2 − 4

Add 4 to both sides

4 = (x + 3)^2

Square root both sides

± 2 = x + 3

Subtract 3 from both sides

x = − 3 ± 2

x intercepts → (x, y) = (− 5, 0)

and

(− 1, 0)

You may if you so choose determine the vertex from this point

x vertex is midway between the x intercepts then by substitution determine y. The method I used later takes less work.

Given

y (x + 3)^2 − 4

x vertex = (-1) ×(+ 3) = − 3

y vertex = − 4

Vertex → (x, y) = (− 3, − 4)

y intercept = + 3^2 − 4 = 5