where does the graph of the function y=tan(x) have asymptotes? at the values of x where cos(x)=0 at the values of x where sin(x)
=0 at the values of x where cos(x) is undefined at the values of x where sin(x) is undefined
2 answers:
Answer: at the values where cos(x) = 0
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.
Justification:
1) tan(x) = sin(x) / cos(x).
2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.
3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.
Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.
You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.
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Answer:
A. -4
Step-by-step explanation:
Slope is rise over run, or y2-y1 / x2-x1.
<u>The slope is A. -4.</u>
Let's see--does that make sense? A slope of four would be pretty steep, and that matches the graph. It also should be negative because the line goes down and to the right, not up and to the right, so -4 makes sense!