Question

f(x)=x^2+28x

Answer 1

question?

Answer 2

How do I find the vertex, the axis of symmetry, and the intcepts?

Answer 3

Parabola
vertex where dy/dx = slope =0 = 2x+28,
x=-14
y = (-14)^2 + 28(-14) = evaluate this

axis of symmetry vertical through this vertex point

intercepts with what? y=0? probably

y=0,
x^2 +28x = 0
use eqn for quadratic

x = [-28 +- sqrt(28^2)]/2

x = 0, -28