Question

Find the vertex and the axis of symmetry for the parabola y= 2x^2 + 8x + 5.

How can I do this without graphing?

Answer 1

recall that the vertex is at x = -b/2a

In this case, that is at x = -8/4 = -2. y(-2) = -3

So, the vertex is at (-2,-3)

You can also determine this by completing the square:

y = 2x^2 + 8x + 5
= 2(x^2 + 4x + 4) - 3
= 2(x+2)^2 - 3

New recall that the vertex of

y = a(x-h)^2 + k

is at (h,k) or (-2,-3) in this case

Answer 2

Wait, so how is y(-2) equal to -3, is the y equal to 1 1/2?

Answer 3

just plug in x = -2

y(-2) = 2(-2)^2 + 8(-2) + 5
= 2*4 - 16 + 5
= 8-16+5
= -3

If y is a function of x, y(2) means evaluate y at x = -2, not multiply y by -2.

Looks like you have some serious reviewing to do.