How do you graph the quadratic function and identify the vertex and axis of symmetry for #y=y=-3x^2+5#?

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Answer 1 · 2017-07-03T02:21:43

Vertex #= (0, +5)#
Axis of symmetry #x=0#

For a quadratic in standard form: #ax^2+bx+c#
The axis of symmetry and vertex will occur where #x=(-b)/(2a)#

In our example: #-3x^2+5 -> a=-3, b=0, c=+5#

#(-b)/(2a) = 0/-6 =0#

Hence the axis of symmetry is #x=0# [i.e. the #y-#axis]

And the y-coordinate of the vertex of the parabola is:
# 0+5=5#

#:.# Vertex #=(0, +5)#

Also, since the coefficient of #x^2 <0 -> y(0)=5# is the absolute maximum of #y#.

The vertex and axis of symmetry can be seen on the graph of #y# below.

graph{-x^2+5 [-11.25, 11.25, -5.63, 5.62]}