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

1 Answer
Jul 19, 2018

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]}