We have this formula for the surface area of a cone
SA = pi*r*s + pi*r^2
The pi*r^2 portion is the flat bottom part of the cone, which is a circle of radius r. The pi*r*s portion is the lateral surface area. This is the triangular part of the cone, so to speak.
"one of the functions represents the surface area of a cone with radius 5 and slant height x". So we can say r = 5 and s = x to go from this
SA = pi*r*s + pi*r^2
to this
SA = pi*5*x + pi*5^2
SA = 5pi*x + 25pi
y = 5pi*x + 25pi
note how this is a linear function in the form y = mx+b
"The other function represents the surface area of a cone with slant height 5 and radius x", meaning s = 5 and r = x
SA = pi*r*s + pi*r^2
SA = pi*x*5 + pi*x^2
SA = 5pi*x + pi*x^2
SA = pi*x^2 + 5pi*x
y = pi*x^2 + 5pi*x
This is a quadratic function due to the x^2 term. This graphs out a parabola.
The graph given to us has f(x) as the parabola (it's the bowl shaped graph) while g(x) is the straight line
Therefore,
f(x) = pi*x^2 + 5pi*x
g(x) = 5pi*x + 25pi
f(x) is the cone with radius x and slant height 5
g(x) is the cone with radius 5 and slant height x
Keep in mind that the slant height of a cone forms the hypotenuse of a right triangle when you do a vertical cross section. If the radius is 5, then the leg of the triangle is 5. The hypotenuse is always longer than either leg, so the hypotenuse x must be larger than 5. A right triangle would not be possible with the hypotenuse x < 5 (you would have to shrink the radius).
Choice D) The function g(x) models the surface area of a cone with radius 5, a slant height of x, and has a domain of x > 5
