Question

How do you find the vertex and intercepts for `y= -4x^2 - 16x -11`?

Answer 1

Vertex: `(-2,5)`

X-Intercepts: `(-(4+sqrt5)/2,0)` and `(-(4-sqrt5)/2,0)`

Refer to the explanation for the process.

Explanation:

Given:

`y=-4x^2-16x-11` is a quadratic equation in standard form:

`ax^2+bx+c`,

where:

`a=-4`, `b=-16`, and `c=-11`.

The axis of symmetry, `x`, is `(-b)/(2a)`.

`x=(-(-16))/(2*-4)`

Simplify.

`x=16/(-8)`

Simplify.

`x=-2`

This is also the `x` value of the vertex.

Vertex: maximum or minimum point of the parabola.

`x=-2`

To determine `y`, substitute `-2` for `x` in the equation and solve.

`y=-4(-2)^2-16(-2)-11`

Simplify.

`y=-4(4)+32-11`

`y=-16+32-11`

`y=5`

The vertex is `(-2,5)`.

Since `a<0`, the vertex is the maximum point and the parabola opens downward.

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

To determine the x-intercepts, substitute `0` for `y` and solve for `x`.

`0=-4x^2-16x-11`

Use the quadratic formula to solve for `x`.

Quadratic formula

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

Plug in the known values from the quadratic equation.

`x=(-(-16)+-sqrt((-16)^2-4*-4*-11))/(2*-4)`

Simplify.

`x=(16+-sqrt(256-176))/(-8)`

Simplify.

`x=(16+-sqrt80)/(-8)`

Prime factorize `80`.

`x=(16+-sqrt((2xx2)xx(2xx2)xx5))/(-8)8`

Simplify.

`x=(16+-4sqrt5)/(-8)`

Simplify.

`x=(4+-sqrt5)/(-2)`

Solutions for `x`.

`x=-(4+sqrt5)/2,` `-(4-sqrt5)/2`

x-intercepts: `(-(4+sqrt5)/2,0)` and `(-(4-sqrt5)/2,0)`

Approximate values of x-intercepts:

`(-0.882,0)` and `(-3.12,0)`

Summary

Vertex: `(-2,5)`

X-Intercepts: `(-(4+sqrt5)/2,0)` and `(-(4-sqrt5)/2,0)`

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=-4x^2-16x-11 [-11.25, 11.25, -5.625, 5.625]}