# How do you find the axis of symmetry, and the maximum or minimum value of the function f(x)=x^2 -2x -15?

Axis of symmetry $x = 1$
Minimum value $= - 16$

#### Explanation:

The parabola opens upward and so this function has a minimum value.
To solve for the minimum value we solve for the vertex.

$y = a {x}^{2} + b x + c$

$y = 1 \cdot {x}^{2} + \left(- 2\right) \cdot x + \left(- 15\right)$

so that $a = 1$ and $b = - 2$ and $c = - 15$

Vertex $\left(h , k\right)$

$h = \frac{- b}{2 a}$

$h = \frac{- \left(- 2\right)}{2 \left(1\right)} = 1$

$k = c - {b}^{2} / \left(4 a\right)$

$k = - 15 - {\left(- 2\right)}^{2} / \left(4 \left(1\right)\right)$

$k = - 15 - 1$

$k = - 16$

Vertex $\left(h , k\right) = \left(1 , - 16\right)$

The minimum value of the function is $f \left(1\right) = - 16$

Kindly see the graph of $f \left(x\right) = {x}^{2} - 2 x - 15$ with the axis of symmetry $x = 1$ dividing the parabola into two equal parts.
graph{(y-x^2+2x+15)(y+1000x-1000)=0[-36,36,-18,18]}

God bless ....I hope the explanation is useful.

May 4, 2016

Axis of symetry $x = 1$
Value of the function $y = - 16$

#### Explanation:

Given -

$y = {x}^{2} - 2 x - 15$

Find Axis of symetry.

$x = \frac{- 2 b}{2 a} = \frac{- \left(- 2\right)}{2 \times 1} = \frac{2}{2} = 1$

Axis of symetry $x = 1$

Maximum of Minimum Values

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 2$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 2 x - 2 = 0$
$x = \frac{2}{2} = 1$

At (x=1): dy/dx=0;(d^2y)/(dx^2)>0
Hence there is a minimum at $x = 1$

Value of the function

$y = {1}^{2} - 2 \left(1\right) - 15$
$y = 1 - 2 - 15 = - 16$