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Graph g(x)=4(x+1)(x+2)(x-1)​

User Leenephi
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Answer:

Explanation:

First, determine the roots (places where the graph crosses the x-axis). They are {-1, -2, 1}. These correspond to the points { (-1, 0}, (-2, 0), (1, 0) }

Next, determine the y-intercept. Let x =0, to determine g(0):

g(0) = 4 (1) (2) (-1), or -8. Thus, the y-intercept is (0, -8).

Next, determine where the graph begins and where it ends. Notice that this is a cubing function, and that when the coefficient of the first term of such a function is positive, the graph begins in the 3rd quadrant and ends in the first.

That is the case for g(x)=4(x+1)(x+2)(x-1)​.

Plot the following: { (-1, 0}, (-2, 0), (1, 0) } These are our zeros. Also, plot the y-intercept, (0, -8).

Starting in the 3rd quadrant, draw a curve upward to (-1, 0), reaching a local maximum at (-1, 0). Continue this curve downward to (0, -8) (the y-intercept). Turning upward again, continue this curve through the second zero, (1, 0).

User Teocomi
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