Factor the following polynomial fully.
1 answer:
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Answer:
A) see attached for a graph. Range: (-∞, 7]
B) asymptotes: x = 1, y = -2, y = -1
C) (x → -∞, y → -2), (x → ∞, y → -1)
Step-by-step explanation:
<h3>Part A</h3>A graphing calculator is useful for graphing the function. We note that the part for x > 1 can be simplified:
This has a vertical asymptote at x=1, and a hole at x=2.
The function for x ≤ 1 is an ordinary exponential function, shifted left 1 unit and down 2 units. Its maximum value of 3^-2 = 7 is found at x=1.
The graph is attached.
The range of the function is (-∞, 7].
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<h3>Part B</h3>As we mentioned in Part A, there is a vertical asymptote at x = 1. This is where the denominator (x-1) is zero.
The exponential function has a horizontal asymptote of y = -2; the rational function has a horizontal asymptote of y = (-x/x) = -1. The horizontal asymptote of the exponential would ordinarily be y=0, but this function has been translated down 2 units.
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<h3>Part C</h3>The end behavior is defined by the horizontal asymptotes:
for x → -∞, y → -2
for x → ∞, y → -1
Take the derivitive
f'(x)=3x^2+12x
find where it equals zero
it equal zero at x=0 and x=-4
find the y values
f(0)=-36
f(-4)=-4
the critical points are (0,-36) and (-4,-4)
make sign chart
evalutat f'(x) at x=-5 and x=-1 and x=1, and see their signs
f'(-5)=(+)
f'(-1)=(-)
f'(1)=(+)
see below attachment
max is where sign changes from (+) to (-)
min is where sign changes from (-) to (+)
so
max at (-4,-4)
min at (0,-36)
none of the options are correct, do you have te right problem?
f'(x)=3x^2+12x
find where it equals zero
it equal zero at x=0 and x=-4
find the y values
f(0)=-36
f(-4)=-4
the critical points are (0,-36) and (-4,-4)
make sign chart
evalutat f'(x) at x=-5 and x=-1 and x=1, and see their signs
f'(-5)=(+)
f'(-1)=(-)
f'(1)=(+)
see below attachment
max is where sign changes from (+) to (-)
min is where sign changes from (-) to (+)
so
max at (-4,-4)
min at (0,-36)
none of the options are correct, do you have te right problem?