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

How can I do this without graphing?

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

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

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.

To find the vertex and axis of symmetry of a parabola in the form y = ax^2 + bx + c, you can use the formula:

x = -b/2a

Let's apply this formula to the given parabola, y = 2x^2 + 8x + 5:

In this case, a = 2 and b = 8.

Using the formula x = -b/2a, we can find the x-coordinate of the vertex:

x = -(8) / (2 * 2)
x = -8 / 4
x = -2

Now, to find the y-coordinate of the vertex, we substitute the x-coordinate (-2) back into the original equation:

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

So the vertex of the parabola is (-2, -3).

Now, let's find the equation for the axis of symmetry. The axis of symmetry is a vertical line that passes through the vertex.

Since the parabola is symmetric about the axis of symmetry, the equation of the axis of symmetry is x = -2.

Therefore, the vertex is (-2, -3) and the axis of symmetry is x = -2.