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

1 Answer
Aug 25, 2017

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]}