Question

Name the vertex, domain and range, and the y-intercept of f(x) = x^2 + 6x + 4

Answer 1

y-intercept is when x = 0, so substitute 0 for x.
domain is the same as it is for all quadratics.
range depends on vertex...
So the vertex:
complete the square:
y = x^2 + 6x + 4 ...

Answer 2

x^2 + 6x + 9 is equal to (x + 3)^2, so add 5 and subtract 5:
y = x^2 + 6x + 4 +5 -5
y = x^2 + 6x + 9 - 5
y = (x+3)^2 -5
vertex form!
Can you get it from here?

Answer 3

i got (-3,-8) is that right?

Answer 4

-3,-5