Find the vertical asymptotes. For each vertical asymptote x = ...

Question

Find the vertical asymptotes. For each vertical asymptote x = a, analyze lim x→a^- f(x) and lim x→a⁺ f(x).

f(x) = (3x⁴ + 3x³ − 36x²) / (x⁴ − 25x² + 144)

Accepted Answer

Welcome back, everyone. Identify any vertical asymptotes of the function F of X. Evaluate limit of F of X X approaches A from the left and from the right for each vertical asymptote X equals A. So first of all, we're given our function F of X, and now F of X is defined as 2 X to the power of 4th plus 6 X cubed minus 20 x 2. And we're dividing that by X to the power of 4 minus 29 X2 + 100. And now we want to identify vertical asymptotes. Let's recall that for a rational function, we are simply setting our denominator equal to 0. And we want to make sure that our numerator is not equal to 0 at the same value of x. Otherwise we have a hole in the function and not a vertical asymptote. So we're writing our numerator, we're setting it to 0 and our denominator, we're setting it not equal to 0. So from here, what we're going to do is analyze the given function and we're going to completely factor it out. For the numerator, we're going to take out 2 X2d. Which leaves us with our second factor X2 + 3 x minus 10. And now in the denominator, well, we can factor out this expression by first of all assuming that x 2 is a variable c, then simplifying it to a quadratic polynomial, finding the factors using the AC method, and this basically gives us X + 2. Multiplied by X minus 2. Multiplied by X + 5 multiplied by X minus 5. Now, let's factor out completely for the numerator, we are going to use the AC method. To factor out X2 + 3 X minus 10. This gives us two factors, X minus 2 multiplied by X + 5, and we're going to rewrite the denominator, X + 2, X minus 2, X + 5, X minus 5. And then we're going to cancel out the removable discontinuities which are holes, right? So X minus 2 is the common factor, then we have X + 5. And we are left with 2 X squared. Divided by X + 2 multiplied by x minus 5. So now, in this case we're going to set each factor to 0, specifically X plus 2 is equal to 0. And therefore the first. Vertical asymptote is going to be X X is equal to -2, which doesn't make the numerator equal to 0, and then X minus 5 is equal to 0, and it follows that X is equal to 5, our second vertical asymptote. The numerator is not equal to 0 at this value, right? So we have our two vertical asymptotes. Let's label them, and now we simply want to evaluate the limits. The first one is limit as X approaches -2 from the left of F of X and we're going to use our. Simplified form to x 2 divided by X + 2 multiplied by x minus 5. We're going to use the direct substitution which gives us 2 multiplied by -2 from the left squared. We're not including from the left because this is going to give us a value that is not equal to 0. But now in the denominator, we have to understand that -2 + 2 gives us 0, but since we're adding -2 from the left plus 2, this gives us 0 from the left, right? Because well essentially X approaches -2 from the left, so X is less than -2. And we're multiplying by -2 minus 5, which is. Negative 7. In the numerator we have a number to multiplied by 4, which is 8. And in the denominator we have 0, but we're approaching 0 from the right now because we have 0 from the left multiplied by a negative number. So in the numerator we have a positive number. In the denominator, it is also positive. Positive divided by positive gives us positive infinity. Now let's analyze the next limit, limit as x approaches -2 from the right. So nothing really changes. We have the same function. We simply have to be careful about the denominator. Now, if we're approaching -2 from the right, we're going to approach 0 from the right. And now we're going to divide a positive number in the numerator by 0 from the left because we're multiplying 0 from the right by -7. So a positive number divided by a negative number is going to give us negative infinity. So we have those two limits and now let's focus on the next 2 limits. Limit as X approaches, our next a value, which is 5. So 5 from the left. And our function is 2 x 2 divided by. X + 2 multiplied by x minus 5. Using the right substitution we're going to get. 2 Multiplied by 5 squared. Divided by 5 + 2, well, basically this is going to be 7. Multiplied by 5 minus 5 gives us 0, right? But we are interested, we're interested whether we are approaching 0 from the left or from the right. If we're approaching 5 from the left, this means that X is less than 5, and therefore X minus 5 is going to be 0 from the left. Once again, in the numerator, we have a positive value. It is equal to 50. Right? In any case, it is a non-zero number, it is simply positive, and we're dividing it by 0. So we're going to get infinity. Since we're dividing it by a negative number, because 7 multiplied by 0 from the left is 0 from the left. A positive number divided by a negative number is a negative number, so we're going to get negative infinity. And now the final limit that we want to evaluate. Is the limit As X approaches 5 from the right. Nothing really changes in this case except from the value of the denominator. In this case, since X is greater than 5, X minus 5 is going to be a limit of 0 from the right. And now we are dividing a positive number by a positive number, so we're going to get positive infinity. And that's it. We have our limits and our vertical asymptotes. Let's label the the final answers. The limits as infinity, negative infinity, negative infinity, and infinity. And we have already labeled the vertical asymptotes X equals -2, and X equals 5. Thank you for watching.

Find the vertical asymptotes. For each vertical asymptote x = a, analyze lim x→a- f(x) and lim x→a+ f(x).f(x) = (3x4 + 3x3 − 36x2) / (x4 − 25x2 + 144)

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Find the vertical asymptotes. For each vertical asymptote x = a, analyze lim x→a^- f(x) and lim x→a⁺ f(x).
f(x) = (3x⁴ + 3x³ − 36x²) / (x⁴ − 25x² + 144)