Question

How do you graph the parabola `y= - x^2 - 6x - 8` using vertex, intercepts and additional points?

Answer 1

See below

Explanation:

Firstly, complete the square to put the equation in vertex form,

`y=-(x+3)^2+1`

This implies that the vertex, or local maximum (since this is a negative quadratic) is `(-3, 1)`. This can be plotted.

The quadratic can also be factorised,

`y=-(x+2)(x+4)`

which tells us that the quadratic has roots of -2 and -4, and crosses the `x axis` at these points.

Finally, we observe that if we plug `x=0` into the original equation, `y=-8`, so this is the `y` intercept.

All of this gives us enough information to sketch the curve:

graph{-x^2-6x-8 [-10, 10, -5, 5]}

Answer 2

First, turn this equation to vertex form:

`y=a(x-h)+k` with `(h,k)` as the `"vertex"`. You can find this by completing the square:

`y=-(x^2+6x+(3)^2-(3)^2)-8`

`y=-(x+3)^2+1`

So the `"vertex"` is at `(-3,1)`

To find the `"zeroes"` also known as `"x-intercept(s)"`, set `y=0` and factor (if it is factorable):

`0=-(x^2+6x+8)`

`0=-(x+4)(x+2)`

`x=-4,-2`

The `"x-intercepts"` are at `(-4,0)` and `(-2,0)`.

You can also use the quadratic formula to solve if it is not factorable (A discriminant that is a perfect square indicates that the equation is factorable):

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(-6)+-sqrt((-6)^2-4*-1*-8))/(2*-1)`

`x=(6+-sqrt(4))/-2`

`x=(6+-2)/-2`

`x=-4,-2`

The `"y-intercept"` is `c` in `ax^2+bx+c`:

The y-intercept here is `(0,-8)`.

To find additional points, plug in values for `x`:

`-(1)^2-6*1-8=>-15=>(1,-15)`

`-(2)^2-6*2-8=>-24=>(2,-24)`

etc.

A graph below is for reference:

graph{-x^2-6x-8 [-12.295, 7.705, -7.76, 2.24]}