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

Question

Nathan Axel · Accepted Answer

Vertex: #(-2,5)#

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

Refer to the explanation for the process.

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]}

Eva Abramson · Answer

undefined To find the vertex of the parabola represented by the equation $y = -4x^2 - 16x - 11$, you first need to find the x-coordinate of the vertex using the formula $-\frac{b}{2a}$, where $a = -4$ and $b = -16$. After finding the x-coordinate, substitute it into the equation to find the corresponding y-coordinate.

To find the x-intercepts, set $y$ to $0$ and solve for $x$. These are the points where the parabola intersects the x-axis.

To find the y-intercept, simply evaluate the equation when $x = 0$. This gives the point where the parabola intersects the y-axis.