Given f(x) = x + 2, describe how setting k = 4 affects the slope and y-intercept of the graph of g compared to the graph of f.
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Simplifying h(x) gives
h(x) = (x² - 3x - 4) / (x + 2)
h(x) = ((x² + 4x + 4) - 4x - 4 - 3x - 4) / (x + 2)
h(x) = ((x + 2)² - 7x - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 14 - 8) / (x + 2)
h(x) = ((x + 2)² - 7 (x + 2) - 22) / (x + 2)
h(x) = (x + 2) - 7 - 22/(x + 2)
h(x) = x - 5 - 22/(x + 2)
An oblique asymptote of h(x) is a linear function p(x) = ax + b such that
In the simplified form of h(x), taking the limit as x gets arbitrarily large, we obviously have -22/(x + 2) converging to 0, while x - 5 approaches either +∞ or -∞. If we let p(x) = x - 5, however, we do have h(x) - p(x) approaching 0. So the oblique asymptote is the line y = x - 5.