How do you find the y intercept, axis of symmetry and the vertex to graph the function #f(x)=x^2-9#?

Question

Serenity Johnson · Accepted Answer

Vertex is at # (0,-9)# , y-intercept: #y= -9 or (0, -9)# ,
axis of symmetry is #x=0# or y-axis.

#f(x) = x^2 -9 = a(x-0)^2 -9 #. Comparing with standard vertex form of equation #f(x)=a(x-h)^2+k ; (h,k)# being vertex, we find here #h=0,k= -9#. So vertex is at #(h,k) or (0,-9)#. putting #x=0# in the equation we find y-intercept as #y= 0^2-9 = -9 or (0, -9)# Axis of symmetry is #x=0# or y-axis. graph{x^2-9 [–40,–40,–20,-20]} [Ans]

Lily Anderson · Answer

undefined To find the y-intercept of the function f(x) = x^2 - 9, substitute x = 0 into the equation:
f(0) = (0)^2 - 9 = -9.
So, the y-intercept is -9.

To find the axis of symmetry of the function, we use the formula for the axis of symmetry of a parabola given by x = -b/2a. For the function f(x) = x^2 - 9, the coefficient of x is 0, so the axis of symmetry is x = 0.

To find the vertex of the parabola, we use the axis of symmetry. Since the axis of symmetry is x = 0, the x-coordinate of the vertex is 0. To find the y-coordinate of the vertex, we substitute x = 0 into the function:
f(0) = (0)^2 - 9 = -9.
So, the vertex of the parabola is (0, -9).