Question

Help find vertex: y=x^2+3x-4

Answer 1

( -3/2 , -25/4 )  is that what you need? Or?

Answer 2

Y=x^2+3x-4

a=1
b=3
c=4

A of sym= negative b over 2a so the axis of symmetry would be 3/2 
plug that into the original equation that's how you find the vertex

y=(3/2)^2+3(3/2)-4
0.75+4.5-4
Vertex= (3/2, 1.25)