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y-intercept. x = 0 Simplify x-intercepts. f(x) = 0 Solve for x. Cubic End behavior. What is that?
Oh the y-intercept is 36
Describe the graph of the function f(x) = x^3 − 11x^2 + 36x − 36. Include the y-intercept, x-intercepts, and the shape of the graph. First of all, the function begins with x to the power of 3: x^3. This is the "cubing function." What does the graph of the cubing function look like? This is important and worth a review of basic graphs. Next: In which of the four quadrants does the graph of y=x^3 begin? As x increases, what direction does the graph take? In which quadrant does the graph end?
Because the given function begins with x^3, it will look somewhat like the graph of y=x^3. Describe the graph of the function f(x) = x^3 − 11x^2 + 36x − 36. Include the y-intercept, x-intercepts, and the shape of the graph. Yes, the y-intercept is found by letting x=0. y = -36, and so that intercept is (0,-36)
tkhunny has already explained how to find the horiz. intercepts: x-intercepts. f(x) = 0 Solve for x.
The zeros are 2, 3, and 6.
So does the graph rise to the right?
Challenge: Can you prove that the zeros are 2, 3 and 6? Hint: Substitute x=6 into the given equation. What value does the function then have?
Yes, the graph rises to the right as x increases, and the graph "ends" in which quadrant?
f(x) = x^3 − 11x^2 + 36x − 36 f(x) = 6^3 - 11(6)^2 + 36(6) - 36 f(x) = 216 - 11(36) + 216 - 36 f(x) = 216 - 396 + 216 - 36 f(x) = -180 + 216 - 36 f(x) = 36 - 36 f(x) = 0
Great. You've just shown that 6 is a zero or root or solution of the original function, set = to 0. Very good. So, you have demonstrated sufficient understanding so that I acdcept your other two roots as being correct. More questions about this problem?
How exactly would I describe the shape of the graph? I know it rises to the right.
The zeros are 2, 3, and 6.
I got it. Thank you. ALSO, are you the one that banned me?
Absolutely not. i did not suspend you.
Glat you've "gotten" this problem! Bye for now.