Question

\[f(x)=4(x+1)^{2}-2\] vertex?

Answer 1

yeah... -1,-2

Answer 2

(-1,-2)

Answer 3

x-intercept of parabola?

Answer 4

f(x)=a(x-h)^2+k gives vertex(h,k)
f(x)=a(x+h)^2+k gives vertex(-h,k)
f(x)=a(x-h)^2-k gives vertex (h,-k)
f(x)=a(x+h)^2-k gives vertex (-h,-k)

Answer 5

to find the x-intercept, set y=0 and solve for x

Answer 6

y = 0 when 1/2 = (x+1)^2

-1 +- sqrt(1/2)