Question

What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola `y= -x^2-8x+10`?

Answer 1

`y=-x^2-8x+10` is the equation of a parabola which because of the negative coefficient of the `x^2` term, we know to open downward (that is it has a maximum instead of a minimum).

The slope of this parabola is
`(dy)/(dx) = -2x-8`
and this slope is equal to zero at the vertex
`-2x-8 =0`
The vertex happens where `x=-4`
`y=-(-4)^2-8(-4)+10 = -16+32+10 =26`
The vertex is at `(-4,58)`
and has a maximum value of `26` at this point.

The axis of symmetry is `x=-4`
(a vertical line through the vertex).

The range of this equation is `(-oo,+26]`

Answer 2

Two other ways to find the vertex of a parabola:

Memorization

The graph of the equation: `y=ax^2+bx+c`,

has vertex at `x=-b/(2a)`

After you use this to find `x`, put that number back into the original equation to find `y` at the vertex.

`y=-x^2-8x+10`, has vertex at `x = - (-8)/(2(-1)) = -8/2 = -4`

The value of `y` when `x=-4` is:

`y=-(-4)^2-8(-4)+10 = -16+32+10=26`.

Complete the Square

Complete the square to write the equation in Vertex Form:

`y = a(x-h)^2 + k` has vertex `(h, k)`.

`y=-x^2-8x+10`

`y=-(x^2+8x color(white)"sssssss")+10`,

`y=-(x^2+8x +16 -16)+10`,

`y=-(x^2+8x +16)-( -16)+10`,

`y=-(x-4)^2+26`, has vertex `(4, 26)`