Question

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation `f(x)=x^2+12x+36`?

Answer 1

graph{x^2 + 12x + 36 [-10, 10, -5, 5]}
`f(x) = x^2 + 12x + 36 = (x + 6)^2`

x-intercepts ->`x = -6` -> Double root. The parabola is tangent to the x axis at x = -6. (Minimum)

x of vertex: x = -6
y of vertex: f(-6) = 36 - 72 + 36 = 0

Axis of symmetry: x = -6