How do you find the vertex, x-intercept, y-intercept, and graph the equation #y=-4x^2+20x-24#?

1 Answer
Dec 17, 2014

The graph is a parabola that opens downward.

Generally, a parabola has the following equation:

# y = ax^2 + bx + c# If a is positive it opens upward. If a is negative it opens downward.

# y= -4x^2 +20x -24#
#a= -4, b = 20, c = -24#

I. Solve for Vertex #( h,k)# Use the formula # h= -b/(2a) =-20/(2(-4)#

# h= 5/2#

To solve for k plug in the value of h to x in the original equation

# y = -4(5/2)^2 +20(5/2) -24#

# y = 1# which is #k#

The vertex is at #(5/2,1)#

II. x and y intercepts

set y = 0 to get the x-intercepts

#0 = -4x^2 + 20x - 24#
divide both sides by -4

# 0 = x^2 -5x + 6#, then factor

#0 = (x-3)(x-2)#

#x= 3 , x =2#

To get the y-intercepts, set # x= 0#

#y = -4(0)^2 +20(0) - 24#

#y= -24#

How to graph?
1. Plot the vertex at # ( 5/2,1)#. This is where opening downward commences.
2. Plot the x intercepts #(3,0), (2,0)# this is where the graph will cross the x-axis.
3. Plot the y-intercept #(0,-24)# This is where the graph crosses the y-axis.
Can anyone help me to create a picture of the graph here. I really don't know how. Thanks.