How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y= -3x^2 + 5#?

1 Answer
Meave60
Sep 28, 2015

Axis of symmetry is #x=0# .
Vertex is #(0,5)#
The graph will be a downward opening parabola.

Explanation:

#y=-3x^x+5# is a quadratic equation in the form #ax^2+bx+c#, where #a=-3, b=0, c=5#.

Axis of Symmetry

The axis of symmetry is determined using the formula #x=(-b)/(2a)#.

#x=(0)/(2*-3)=(0)/-6=0#

The axis of symmetry is #x=0#.

Vertex

The vertex is the maximum or minimum point on the parabola. In this case, since the coefficient of #x^2# is #-3#, the parabola will open downward and the vertex will be the maximum point.

The #x# value of the vertex is #0# from the axis of symmetry.

To find the value of #y# for the vertex, substitute #0# for #x# in the equation and solve for #y#.

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

#y=0+5=5#

The vertex is #(0,5)#.

Determine several points on both sides of the axis of symmetry.

#x=-2#, #y=-7#
#x=-1#, #y=2#
#x=0#, #y=5# (vertex)
#x=1#, #y=2#
#x=2,# #y=-7#

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

graph{y=-3x^2+5 [-16.02, 16, -8.01, 8.01]}

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1262 views around the world
You can reuse this answer
Creative Commons License