Question

Describe the graph of the function f(x) = x3 − 11x2 + 36x − 36. Include the y-intercept, x-intercepts, and the shape of the graph.

Answer 1

y-intercept: (0, -36); x-intercepts: (2,0), (3,0), (6,0)

Explanation:

Hint: Please use " ^ " to denote exponentiation: f(x) = x^3 − 11x^2 + 36x − 36

I guessed that a root of this function is x = 6, with the implication that (x-6) is a factor of f(x) = x^3 − 11x^2 + 36x − 36.

This can be proven using synthetic division. The coefficients of the quotient are 2, 3 and 6. Thus, the x-intercepts are (2,0), (3,0) and (6,0). Letting x = 0 leaves us with y= -36, so the y-intercept is (0, -36)

The graph begins in Quadrant III, increasing as x increases. It enters Quadrant I briefly and then it reverses direction and heads downward a bit, and finally turning up and increasing indefinitely in Quadrant I.

Answer 2

Explanation:

y intercept occurs when x = 0

here it is -36.

As the last number is -36, 6 may be a x intercept here

f(x) = 6^3 - 11*36 - 36*6 - 16 - 0 so x = 6 is a x-intercept ( zero) of this graph.

Dividing the function by (x - 6) gives x^2 - 5x + 6

which = (x - 2)(x - 3) so the other 2 zeroes are 2 and 3.

The coefficient of x^3 is positive so the curve will rise from the left passing through y = - 36 and forming 2 turning points, cutting the x axis at x = 2, 3 and 6 and will rise to the right. There will be a local maximum between x = 2 and x = 3 and a local minimum between x = 3 and x = 6.