How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #f(x)=9-(x+3)^2#?

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Answer 1 · 2015-06-09T04:56:48

f(x) = -(x + 3)^2 + 9

f(x) is expressed in vertex form.

x of vertex: x = -3

y of vertex y = +9

x of axis of symmetry = x of vertex = -3

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

(x + 3)^2 = 9 -> (x + 3) = +-3

x = -3 + 3 = 0
x = -3 - 3 = -6.

To find y-intercept, make x = 0
f(x) = 9 - 9 = 0