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1 answer:
Answer: (A) vertical asymptote: x = 2, horizontal asymptote: y = 1
<u>Step-by-step explanation:</u>
<u>Vertical Asymptote</u> is the restriction on the x-value. The denominator cannot be zero, so x - 2 ≠ 0 ⇒ x ≠ 2
The restricted value on x is when x = 2 <em>which is the vertical asymptote</em>
<u>Horizontal Asymptote</u> (H.A.) is the restriction on the y-value. This is a comparison of the numerator (n) and denominator (m). There are 3 rules that will help you:
- n > m No H.A. (use long division to find slant asymptote)
- n = m H.A. is the coefficient of n divided by coefficient of m
- n < m H.A. is 0
In the given problem, n < m so y = 0, however there is also a vertical shift of up 1 so the H.A. also shifts up. This results in H.A. of y = 1
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Answer:
See attached for graphs
g(x) -- domain: -∞ < x < ∞; range: 0 < y < ∞
g^-1(x) -- domain: 0 < x < ∞; range: -∞ < y < ∞
Step-by-step explanation:
g(x) is an exponential decay function. Its base is 1/3, so each increase of 1 unit in x will multiply the y-value by a factor of 1/3. The graph will rapidly approach its horizontal asymptote of y=0 as x gets large. The y-intercept is (0, 1). Just as y gets smaller as x increases, so it gets larger as x decreases. Each decrease of x by 1 unit causes the y-value to be multiplied by 3.
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The graph of g^-1(x) is the graph of g(x) reflected across the line y=x. That is, each coordinate pair (x, y) on the graph of g(x) becomes a point (y, x) on the graph of the inverse function. In order to graph g^-1(x), you don't need to write down the function, you only need to know the relationship between the graphs.
Just as x- and y- are interchanged on the graph, so the domain, range, and intercepts are interchanged. g^-1(x) will have a vertical asymptote of x=0, and an x-intercept of (1, 0). The domain of g^-1(x) is the range of g(x): 0 < x < ∞; and the range of g^-1(x) is the domain of g(x): -∞ < y < ∞.
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The attached graph shows g(x) in red and g^-1(x) in blue. As you can see, we created the graph simply by interchanging x and y. The line y=x is shown for reference, so you can see that each curve is a reflection of the other across that line.
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<em>Additional comment</em>
The explicit expression for g^-1(x) can be found by solving for y:
x = g(y)
If you're familiar with the log function, you know it has an x-intercept of 1 and a vertical asymptote at x=0. The base of the log function is simply a vertical scale factor. The minus sign reflects it across the x-axis.
