Question

What is the axis of symmetry and vertex for the graph `y=x^2+3x-5`?

Answer 1

Vertex at `(-3/2,-29/4)`. Axis of symmetry is `x= -3/2`

Explanation:

`y=x^2+3x-5 = (x+3/2)^2-9/4-5=(x+3/2)^2-29/4:.`Comparing with general vertex form of equation `y=a(x-h)^2+k` we get vertex at `(h,k) or (-3/2,-29/4)`.Axis of symmetry is `x= -3/2` graph{x^2+3x-5 [-20, 20, -10, 10]} [Ans]