Question

Please help!

Find the vertex of f(x)=x^2+2x-3
Find the vertex of f(x)=x^2-3x+1

Answer 1

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

Answer 2

Thank you!