Question

How do you graph `x ^ { 2} = 2x + 99`?

Answer 1

Refer to the explanation.

Explanation:

Graph:

`x^2=2x+99`

Move all terms to the left-hand side.

`x^2-2x-99=0`

Switch sides.

`0=x^2-2x-99`

The standard form for a quadratic equation is:

`y=ax^2+bx+c`,

where:

`a=1`, `b=-2`, `c=-99`

You need the vertex, y-intercept, and x-intercepts to graph a quadratic equation.

The vertex is the minimum or maximum point of a parabola. The x-coordinate is determined using the formula for the axis of symmetry, and the y-coordinate is determined by substituting the value of `x` into the equation and solving for `y`.

Axis of symmetry:

`x=(-b)/(2a)`

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

`x=2/2`

`x=1`

The x-coordinate of the vertex is `1`.

Substitute `y` for `0`, and `1` for `x`. Solve for `y`.

`y=(1)^2-2(1)-99`

`y=1-2-99`

`y=-100`

The vertex is `(1,-100)`

Y-intercept: value of `y` when `x=0`. Substitute `0` for `x` and solve for `y`.

`y=0^2-2(0)-99`

`y=-99`

The y-intercept is `(0,-99)`

X-intercepts: values of `x` when `y=0`.

Substitute `0` for `y` and solve for `x`.

`0=x^2-2x-99`

Find two numbers that when added equal `-2` and when multiplied equal `-99`. The numbers `-9` and `11` meet the requirements.

`0=(x+9)(x-11)`

Set each binomial equal to `0` and solve for `x`.

`x+9=0`

`x=-9`

`x-11=0`

`x=11`

The x-intercepts are `(-9,0)` and `(11,0)`.

Plot the vertex, y-intercept, and x-intercepts. Sketch a parabola through the points. Do not connect the dots

graph{y=x^2-2x-99 [-18.2, 17.83, -12.61, 5.4]}