How do you find the vertex and axis of symmetry, and then graph the parabola given by: y=x^(2)-2x-15?

Aug 18, 2017

The axis of symmetry is $1$.

The vertex is $\left(1 , - 16\right)$

The x-intercepts are $\left(5 , 0\right)$ and $\left(- 3 , 0\right)$.

Refer to the explanation for the process.

Explanation:

Given:

$y = {x}^{2} - 2 x - 15$ is a quadratic equation in standard form:

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

where:

$a = 1$, $b = - 2$, and $c = - 15$

The formula for the axis of symmetry and the $x$ value of the vertex is:

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

Plug in the known values.

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

Simplify.

$x = \frac{2}{2} = 1$

To find the $y$ value of the vertex, substitute $1$ for $x$ in the equation.

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

Simplify.

$y = 1 - 2 - 15$

$y = - 16$

The vertex is $\left(1 , - 16\right)$. Since $a > 0$, the vertex is the minimum point and the parabola opens upward.

We also need the x-intercepts, which are the values of $x$ when $y = 0$. Substitute $0$ for $y$ and solve for $x$ by factoring.

$0 = {x}^{2} - 2 x - 15$

Find two numbers that when added equal $- 2$, and when multiplied equal $- 15$. The numbers $- 5$ and $3$ meet the requirements.

$0 = \left(x - 5\right) \left(x + 3\right)$

$0 = x - 5$

$5 = x$

$0 = x + 3$

$- 3 = x$

$x = 5 , - 3$

The x=intercepts are $\left(5 , 0\right)$ and $\left(- 3 , 0\right)$.

Summary

The axis of symmetry is $1$.

The vertex is $\left(1 , - 16\right)$

The x-intercepts are $\left(5 , 0\right)$ and $\left(- 3 , 0\right)$.

Now plot the vertex and the x-intercepts. Sketch a parabola through the points. Do not connect the dots.

graph{x^2-2x-15 [-14.82, 17.2, -16.08, -0.06]}