Question

What is the y-value of the vertex of 4x^2 + 8x-8?

Answer 1

To find the y-value of the vertex of a quadratic function in the form of ax^2 + bx + c, where a, b, and c are constants, we can use the formula:

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

where (h, k) is the vertex of the parabola.

For the given quadratic function 4x^2 + 8x - 8, we can rewrite it in vertex form:

4x^2 + 8x - 8 = 4(x^2 + 2x) - 8
4(x^2 + 2x + 1) - 8 = 4(x + 1)^2 - 8

Comparing this to the vertex form equation y = a(x - h)^2 + k, we can see that the vertex of the parabola is at (-1, -8).

Therefore, the y-value of the vertex of the given function is -8.

Answer 2

well the Bot used calculus for the last one so I will this time.
when is dy/dx = slope of parabola = 0
y = 4x^2 + 8x-8
dy/dx = 0 = 8 x + 8
so when x = -1
then y = 4 - 8 - 8 = -12

Answer 3

Thank you for pointing that out.
Using calculus to find the y-value of the vertex is also a valid method.

By setting the derivative dy/dx of the function 4x^2 + 8x - 8 equal to 0, we can find the x-coordinate of the vertex.

Solving for x:
dy/dx = 8x + 8 = 0
8x = -8
x = -1

Substitute x = -1 back into the original equation to find the y-value at the vertex:
y = 4(-1)^2 + 8(-1) - 8
y = 4(1) - 8 - 8
y = 4 - 8 - 8
y = -12

Therefore, the y-value of the vertex of the function is -12.