How to graph a parabola #y=-3(x-2)^2 + 5#?

1 Answer
Jun 29, 2018

Refer to the explanation.

Explanation:

Graph:

#y=-3(x-2)^2+5# is a quadratic equation in vertex form:

#y=a(x-h)^2+k#,

where:

#h# is the axis of symmetry and #(h,k)# is the vertex.

In order to graph a parabola, you need the vertex, the y-intercept, x-intercepts, and one or more additional points.

Vertex: maximum or minimum point of the parabola. Since #a<0#, the vertex is the maximum point and the parabola opens downward.

The vertex is #(2,5)#

Y-intercept: value of #y# when #x=0#

Substitute #0# for #x#.

#y=-3(0-2)^2+5#

#y=-3(-2)^2+5#

#y=-3(4)+5#

#y=-7#

The y-intercept is #(0,-7)#.

X-intercepts (Roots): values for #x# when #y=0#

Substitute #0# for #y# and solve for #x#.

#0=-3(x-2)^2+5#

Subtract #5# from both sides.

#-5=-3(x-2)^2#

Divide both sides by #-3#.

#5/3=(x-2)^2# #larr# Two negatives make a positive.

Take the square root of both sides.

#+-sqrt(-5/3)=x-2#

Use the rule #sqrt(a/b)=(sqrta)/(sqrtb)#

#+-(sqrt5)/(sqrt3)=x-2#

Add #2# to both sides.

#2+-(sqrt5)/(sqrt3)=x#

Switch sides.

#x=2+-(sqrt5)/(sqrt3)#

#x=2+(sqrt5)/(sqrt3)#, #2-(sqrt5)/(sqrt3)#

#x=~~3.291, 0.709#

The x-intercepts are: #(2+(sqrt5)/(sqrt3),0)#, #(2-(sqrt5)/(sqrt3))#

Approximate x-intercepts: #(~~3.291,0)#, #(0.709,0)#

Additional point:

Substitute #4# for #x# and solve for #y#.

#y=-3(4-2)^2+5#

#y=-3(2)^2+5#

#y=-3(4)+5#

#y=-12+5#

#y=-7#

Additional point: #(4,-7)#

Summary:

Vertex: #(2,5)#

Y-intercept: #(0,-7)#

X-intercepts: #(2+(sqrt5)/(sqrt3),0)#, #(2-(sqrt5)/(sqrt3))#

Approximate x-intercepts: #(~~3.291,0)#, #(0.709,0)#

Additional point: #(4,-7)#

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

graph{y=-3(x-2)^2+5 [-10, 10, -5, 5]}