What is the axis of symmetry and vertex for the graph #y=-x^2-3x+2#?

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Answer 1 · 2017-03-19T11:14:30

The axis of symmetry is #x=-3/2#
The vertex is #=(-3/2,17/4)#

We use

#a^2-2ab+b^2=(a-b)^2#

We complete the square and factorise in order to find the vertex form.

#y=-x^2-3x+2#

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

#y=-(x^2+3x+9/4)+2+9/4#

#y=-(x+3/2)^2+17/4#

This is the vertex form of the equation.

The axis of symmetry is #x=-3/2#

The vertex is #=(-3/2,17/4)#

graph{(y+(x+3/2)^2-17/4)((x+3/2)^2+(y-17/4)^2-0.02)(y-1000(x+3/2))=0 [-11.25, 11.25, -5.625, 5.625]}