Curve sketching practice question 17
Find the slant asymptote of the graph of .
The top has degree 2 and the bottom degree 1. Which operation splits into a line plus a part that tends to 0?
Divide: write and match coefficients, or use long division.
Then , and the remainder term tends to 0 as : the quotient is the asymptote.
An equation of a line, y = mx + b. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §7.1CLP-1 §3.6Key resultsBeyond the course text
Worked solution
Strategy
Divide the top by the bottom. The quotient is a line, and the remainder over x − 1 tends to 0, so the graph approaches that line at both ends.
Working
Long division: 2x² ÷ x = 2x; subtracting 2x² − 2x leaves 5x − 1; 5x ÷ x = 5; subtracting 5x − 5 leaves 4.
Divide through by x − 1.
The graph approaches the line y = 2x + 5 at both ends: from above on the right, from below on the left.
Answer
The answer is y equals 2 x plus 5.
Check
At x = 101, f(x) = 20,704/100 = 207.04, just 0.04 above 2x + 5 = 207.
Where this goes wrong
Stopping at y = 2x from 2x²/x, or reading off y = 2 from the leading coefficients. Only the full quotient, 2x + 5, is the line the graph approaches.
The idea behind it
The curve sketching lesson, a figure to move, in a few steps
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