This function is a special form of a quadratic:
#color(red)(x)^2 - color(blue)(y)^2 = (color(red)(x) + color(blue)(y))(color(red)(x) - color(blue)(y))#
Let:
#color(red)(x)^2 = color(red)(9)color(red)(x)^2#; then #color(red)(x) = color(red)(3x)#
#color(blue)(y)^2 = color(blue)(64)#; then #color(blue)(x) = color(blue)(8)#
Substituting this into the function gives:
#f(x) = (color(red)(3x) + color(blue)(8))(color(red)(3x) - color(blue)(8))#
To find the #x#-intercepts we set the right side of the function to #0# and solve each term for #0#:
#(color(red)(3x) + color(blue)(8))(color(red)(3x) - color(blue)(8)) = 0#
Solution 1:
#3x + 8 = 0#
#3x + 8 - color(red)(8) = 0 - color(red)(8)#
#3x + 0 = -8#
#3x = -8#
#(3x)/color(red)(3) = -8/color(red)(3)#
#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -8/3#
#x = -8/3#
Solution 2:
#3x - 8 = 0#
#3x - 8 + color(red)(8) = 0 + color(red)(8)#
#3x - 0 = 8#
#3x = 8#
#(3x)/color(red)(3) = 8/color(red)(3)#
#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 8/3#
#x = 8/3#
The x-intercepts are:
#x = -8/3# and #x = -8/3#
Or
#(-8/3, 0)# and #(8/3, 0)#
