First, we can set #x = # to find the #y#-intercept:
#y = x^2 - 9# becomes:
#y = 0^2 - 9#
#y = 0 - 9#
#y = -9# so the #y#-intercept is #(0, -9)# as one of the points on the graph.
Next we can find the #0#s or the #x# intercepts.
The right side of the equation is the special form of a quadratic:
#(color(red)(a) + color(blue)(b))(color(red)(a) - color(blue)(b)) = color(red)(a)^2 - color(blue)(b)^2#
If we let #color(red)(a)^2 = x^2# then #color(red)(a) = x#
If we let #color(blue)(b)^2 = 9# then #color(blue)(b) = 3#
Substituting this into the formula gives:
#y = (color(red)(x) + color(blue)(3))(color(red)(x) - color(blue)(3))#
Solving each of these terms for #0# gives us the #x#-intercepts or #0#s.
#color(red)(x) + color(blue)(3) = 0#
#color(red)(x) + color(blue)(3) - 3 = 0 - 3#
#color(red)(x) = -3#
And
#color(red)(x) - color(blue)(3) = 0#
#color(red)(x) - color(blue)(3) + 3 = 0 + 3#
#color(red)(x) = 3#
The #0#'s are: #(color(red)(-3), 0)# and #(color(red)(3), 0)#
Plotting these points and drawing the parabola gives:
graph{y = x^2 - 9 [-6, 6 -15, 15]}
