How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y=x^2+3x-4#?
y = x^2 + 3x - 4
x of vertex = x of axis of symmetry = x = (-b/2a) = -3/2
y of vertex: y = f(-3/2) = 9/4 - 9/2 - 4 = -25/4
To get y-intercept, make x = 0 -> y = -4.
To get x-intercept, solve y = 0.
Since a + b + c = 0, use shortcut. One real root (x-intercepts) is (1) and the other is (c/a = -4).