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# Tutor profile: Sandy R.

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Sandy R.
Math Educator with 23 years of Experience
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## Questions

### Subject:Pre-Calculus

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Question:

I am supposed to find the vertical asymptotes of the function $$f(x)=\frac{x^2-x-2}{x+1}$$. I set the bottom x + 1 = 0 and found x = -1 as the vertical asymptote. However, my answer key in my book says there are no vertical asymptotes. I don't understand why? Can you help?

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Sandy R.

You are thinking correctly, but you also need to look at the numerator and see if you can simplify the overall function. Notice you can factor the numerator. $$x^2-x-2=(x-2)(x+1)$$ So you can rewrite your function as $$f(x)=\frac {(x-2)(x+1)}{x+1}$$ You can simplify the function by canceling the x+1 factors on top and bottom. This means that x = -1 is a hole in the graph instead of a vertical asymptote. Any factors that cancel on the top and bottom result in a hole in the graph. Any factors that remain in the denominator of your function will result in vertical asymptotes. Since no factors remain on the bottom of your function there is no vertical asymptotes.

### Subject:Trigonometry

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Question:

I need help finding all the solutions of the trig equation $$cosx + 1 = sinx$$ on the interval $$[0,2\pi]$$ Do I solve each side of the equation separately by setting them equal to zero?

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Sandy R.

To start to solve this equation you need to first get the equation in terms of a single trigonometric function. Once you have all the terms in the same trig function you can then get all the terms on one side and set the whole equation equal to zero. If you square each side of the equation you can use the Pythagorean Identity. $$(cosx-1)^2=sin^2x$$ $$(cosx-1)(cosx-1)=sin^2x$$ $$cos^2x+2cosx+1=sin^2x$$ Recall the Pythagorean Identity $$sin^2x+cos^2x=1$$. So you can rewrite $$sin^2x$$ as $$1-cos^2x$$ So we will have $$cos^2x+2cosx+1=1-cos^2x$$ Now get all your terms on the left side and you can set your equation equal to zero and solve from here.

### Subject:Calculus

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Question:

I need help finding the $$\int x \sqrt{2x-1} dx$$ I think I need to use a substitution, but I don't know what to make "u"?

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Sandy R.

Yes, you are on the right track. This does require a substitution. However, there is an extra step where you are going to need to change your variables. Let u =2x-1 the part underneath the square root symbol. Then your derivative will be du = 2 dx. Solve for dx to get $$\frac{du}{2}=dx$$ You probably noticed this doesn't match the part of the equation on the outside of the square root. We need to match the "x" . This is why we need to change our variable. Go back to your substitution. We have u = 2x-1 We need to solve this for x. We have u = 2x - 1 u + 1 = 2x $$\frac{(u+1)}{2}=x$$ You can now substitute back into your expression $$\int x \sqrt{2x-1} dx$$ We will have $$\int\frac{u+1}{2} u^\frac{1}{2}(\frac{du}{2})$$ From this point we can multiply and simplify the expression and then integrate as normal.

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