Question

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

Answer 1

The axis of symmetry is `1`.

The vertex is `(1,-16)`

The x-intercepts are `(5,0)` and `(-3,0)`.

Refer to the explanation for the process.

Explanation:

Given:

`y=x^2-2x-15` is a quadratic equation in standard form:

`y=ax^2+bx+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=(-b)/(2a)`

Plug in the known values.

`x=(-(-2))/(2*1)`

Simplify.

`x=2/2=1`

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

`y=1xx1^2-2(1)-15`

Simplify.

`y=1-2-15`

`y=-16`

The vertex is `(1,-16)`. 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^2-2x-15`

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

`0=(x-5)(x+3)`

`0=x-5`

`5=x`

`0=x+3`

`-3=x`

`x=5,-3`

The x=intercepts are `(5,0)` and `(-3,0)`.

Summary

The axis of symmetry is `1`.

The vertex is `(1,-16)`

The x-intercepts are `(5,0)` and `(-3,0)`.

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]}