Question

How do you find the vertex and the intercepts for `e(x) = -x² - 8x - 16`?

Answer 1

The vertex is `(-4,0)`.

There are no x-intercepts.

Explanation:

Given:

`e(x)=-x^2-8x-16`,

where:

`a=-1`, `b=-8`, `c=-16`

Vertex: minimum or maximum point of the parabola

The x-coordinate of the vertex is found using the formula:

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

`x=(-(-8))/(2*-1)`

`x=8/(-2)`

`x=-8/2`

`x=-4`

Substitute `y` for `e(x)`. The y-coordinate of the vertex is found by substituting `-4` for `x` and solving for `y`.

`y=-(-4)^2-8(-4)-16`

`y=-16+32-16`

`y=0`

The vertex is `(-4,0)`.

Use the discriminant (D) from the quadratic formula to determine whether there are any x-intercepts.

`"D"=sqrt(b^2-4ac)`

Plug in the known values.

`"D"=sqrt((-8)^2-4*-1*-16)`

`"D"=sqrt(-128)`

Since the discriminant is the square root of a negative number, there are no real solutions (x-intercepts).

You could determine the y-intercept and the point opposite the y-intercept to help graph the parabola.

Y-intercept: value of `y` when `x=0`

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

`y=-(0)^2-8(0)-16`

`y=-16`

The y-intercept is `(0,-16)`

Opposite point

`x=-8`

Substitute `-8` for `x` and solve for `y`.

`y=-(8^2)-8(-8)-16`

`y=-16`

The opposite point is `(-8,-16)`

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

graph{-x^2-8x-16 [-10, 10, -5, 5]}