SOLUTION: Let f(x) = -3x(x+2)^2 (x-4)^3
Perform the Leading Coefficient Test to determine the end behavior of the function. List a^n and n and explain how they lead to your conclusion.
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-> SOLUTION: Let f(x) = -3x(x+2)^2 (x-4)^3
Perform the Leading Coefficient Test to determine the end behavior of the function. List a^n and n and explain how they lead to your conclusion.
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Question 926832: Let f(x) = -3x(x+2)^2 (x-4)^3
Perform the Leading Coefficient Test to determine the end behavior of the function. List a^n and n and explain how they lead to your conclusion.
(Identify all zeros) Found 2 solutions by ewatrrr, Theo:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! f(x) = -3x(x+2)^2 (x-4)^3
for x > 4, (x-4)^3 > 0 and therefore, end behavior for f(x) = -inf, due to the negative Leading Coefficient
You can put this solution on YOUR website! your equation is:
f(x) = -3x(x+2)^2 (x-4)^3
(x+2)^2 will have a leading term of x^2
(x-4)^3 will have a leading term of x^3
x * x^2 * x^3 = x^6
your leading exponent is x^6
your leading coefficient will be negative because you have 2 leading coefficients that are positive times 1 leading coefficient that is negative to get a negative leading coefficient because a negative times 2 positives is equal to a negative.
your leading term have an even exponent and your leading coefficient is negative so the equation will fall to the left and fall to the right.
your zeroes are x = 0, x = -2, x = 4
the graph of the equation will touch the x-axis at x = -2 and cross the x-axis at x = 0 and x = 4
this is because the exponent of the (x+2)^2 term is even and the exponent of the (x-4)^3 term and the x term are both odd.
the graph of your equation looks like this:
here's a reference that can help you understand just what went on.
