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

Apr 28, 2015

$y = - {x}^{2} - 8 x + 10$ 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
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x - 8$
and this slope is equal to zero at the vertex
$- 2 x - 8 = 0$
The vertex happens where $x = - 4$
$y = - {\left(- 4\right)}^{2} - 8 \left(- 4\right) + 10 = - 16 + 32 + 10 = 26$
The vertex is at $\left(- 4 , 58\right)$
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 $\left(- \infty , + 26\right]$

Apr 28, 2015

Two other ways to find the vertex of a parabola:

Memorization

The graph of the equation: $y = a {x}^{2} + b x + c$,

has vertex at $x = - \frac{b}{2 a}$

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

$y = - {x}^{2} - 8 x + 10$, has vertex at $x = - \frac{- 8}{2 \left(- 1\right)} = - \frac{8}{2} = - 4$

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

$y = - {\left(- 4\right)}^{2} - 8 \left(- 4\right) + 10 = - 16 + 32 + 10 = 26$.

Complete the Square

Complete the square to write the equation in Vertex Form:

$y = a {\left(x - h\right)}^{2} + k$ has vertex $\left(h , k\right)$.

$y = - {x}^{2} - 8 x + 10$

$y = - \left({x}^{2} + 8 x \textcolor{w h i t e}{\text{sssssss}}\right) + 10$,

$y = - \left({x}^{2} + 8 x + 16 - 16\right) + 10$,

$y = - \left({x}^{2} + 8 x + 16\right) - \left(- 16\right) + 10$,

$y = - {\left(x - 4\right)}^{2} + 26$, has vertex $\left(4 , 26\right)$