Plug-in values into function to get point of function
Find the roots of x2 + 5x + 6
x2 + 5x + 6 = (x + 3)(x + 2)
(x + 3)(x + 2) = 0
x = −3, −2
Find the roots of 4x2 + 8x + 2
4x2 + 8x + 2 = 2(2x2 + 4x + 1) = 2(2x + 1)2
Two identical roots, x = −[1/2], −[1/2]
Find the asymptote(s) of f(x) = [1/(4x2 + 8x + 2)]
We already found the roots on the previous problems. The asymptote is where the function is undefined. Here, it's when the denominator is equal to zero. We know that the denominator is equal to zero when x = −[1/2]. The function goes to positive infinity when approaching the root (−[1/2]) from the positive direction. The function also goes to positive infinity when approaching the root from the negative direction (The degree of the polynomial in the denominator is even).
Graph y = [1/(4x2 + 8x + 2)]
We already determined the asymptote behavior, what about the edge behavior? As x gets very large and positve, y approaches zero. As x gets very large and negative, y also approaches zero from the positive side.
Find the asymptote(s) of f(x) = [1/x]
The function is undefined at x = 0. The function goes to positive infinity when approaching zero from the positive direction. The function goes to negative infinity when approaching zero from the negative direction (Remember, the polynomial in the denominator is of odd degree).
Find the asymptote(s) of f(x) = tanx
tanx = [sinx/cosx]. The function is undefined when cosx = 0. cosx = 0 when x = ..., −[(π)/2], [(π)/2], [(3π)/2], [(5π)/2], ...
There are an infinite number of asymptotes for f(x) = tanx and those asymptotes are at odd integer multiples of [(π)/2]
This is similar to the graph of y = [1/x], but it is shifted 5 to the left.
Graph y = [1/(x2 − 4)]
Even degree polynomial in the denominator. As x gets very large in either direction, y approaches zero. There are two asymptotes here, -2 and 2. That gives us 4 different cases for asymptotic behavior. Approaching -2 from the left, Approaching -2 from the right, approaching 2 from the left, and approaching 2 from the right. Plugging in values to test for these cases.
f(−2.001) ≈ 250
f(−1.999) ≈ −250
f(1.999) ≈ −250
f(2.001) ≈ 250
Graph y = [x/(x2 −4)]
Asymptotes are at same locations as previous problem, so we can find the asymptote behavior in a similar fashion
f(−2.001) ≈ −500
f(−1.999) ≈ 500
f(1.999) ≈ −500
f(2.001) ≈ 500
Graph f(x) = [1/lnx]
Remember the properties of lnx. Domain (0, ∞), ln1 = 0, and lnx is an increasing function. As x gets larger, f(x) approaches zero and f(x) has an asymptote at x = 1.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
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