Question

identify vertex,axis of symmetry,,max or min,and domain and range of y=(x+8)^2-12

Answer 1

notice that your equation is already in "vertex form" -> http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

can you spot the vertex? and axis of symmetry?

Answer 2

no..help me plzz!!!!

Answer 3

\(\bf y=(x+\color{red}{8})^2\color{red}{-12}\qquad \qquad y = (x-\color{red}{h})^2\color{red}{+k}\\ \quad \\ \quad \\
vertex \to (\color{red}{h, k})\)

Answer 4

thanx alot ,it was a great deal of help

Answer 5

keep in mind that the form is \(\bf (x+h)^2+k\) thus \(\bf y=(x+8)^2-12\implies y=(x-(-8))^2-12\)

Answer 6

so your vertex is at (-8, -12)

Answer 7

hmm I mean....
the vertex form is \(\bf (x-h)^2+k\)
and
\(\bf y=(x+8)^2-12\implies y=(x-(-8))^2-12\)

thus the vertex is at (-8, -12)

Answer 8

and theother parts???

Answer 9

the squared binomial is positive, that means the parabola goes upwards, or

can you see the axis of symmetry?
that is, where the side on the left, begins to look like the side on the right in the parabola?

Answer 10

if I were to "cut in 2 equal parts" that graph, I'd have to draw a line through