x^2-16, whats the Axis of symetry? Vertex? X/Y-int?

Question

plz help >.<

 

Answer 1

Hi Kit,

 

This is a quadratic equation. The coefficients are a=1, b=0 (think of it as 0*x in the equation) and c=-16

The axis of symmetry is the line x=−b/2a

b = 0 so the axis of symmetry is x = 0

The y value at the axis of symmetry is the y-coordinate of the vertex, which is also the y-intercept.

plug in x=0 to find that

0^2 - 16 = -16

Vertex is at (0, -16) 

y -intercept = -16

x intercepts are the roots of the equation so if solve for x, we can find them

x^2 - 16 = 0

x^2 = 16

x = {-4, 4}