Answer:
1a: f(x) = x³ -6x² +9x -4
1b: no breaks; the function is defined everywhere
2a: (-∞, 5]
2b: y = -3; y = -1
2c: x → -∞, y → -3; x → ∞, y → -1
Step-by-step explanation:
<h3>1A:</h3>The factored form of f(x) is ...
f(x) = (x -4)(x -1)²
f(x) = (x -4)(x² -2x +1) = x³ -2x² +x -4x² +8x -4
f(x) = x³ -6x² +9x -4
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<h3>1B:</h3>The function h(x) is defined for all values of x to be either f(x) or g(x). The function f(x) is a polynomial function, so has no breaks in its domain. The function g(x) is defined for all values of x, so has no breaks in its domain.
There are no breaks in the domain of h(x).
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<h3>2A:</h3>See the attachment for a graph.
-∞ < y ≤ 5 . . . . range of f(x)
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<h3>2B:</h3>The asymptote as x → -∞ is the asymptote of the exponential function. The exponential term has an asymptote of y=0, but shifting it down 3 units means the asymptote is 0 -3:
x → -∞, y → -3 . . . . asymptote is y = -3
The asymptote as x → ∞ is the asymptote of the rational expression. That is the ratio of the leading terms: -x²/x² = -1.
x → ∞, y → -1 . . . . asymptote is y = -1
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<h3>2C:</h3>See part B.
