the vertex is on the axis of symmetry
... x = -b / 2a
f(x)=x^2+2x-3
... x = -2 / 2*1 = -1
f(x)=x^2-3x+1
... x = - -3 / 2*1 = 3/2
plug the x's into their functions to find the y's for the vertex coordinates
Please help!
Find the vertex of f(x)=x^2+2x-3
Find the vertex of f(x)=x^2-3x+1
2 answers
Thank you!