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# Tutor profile: Victoria B.

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Victoria B.
Senior Math Major with Several Years of Tutoring Experience
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## Questions

### Subject:Pre-Calculus

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Question:

Find the holes, $$x$$-intercepts, $$y$$-intercepts, and asymptotes of the function $$f(x)=\frac{x^3-x^2-2x}{3x^2+x-14}$$

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Victoria B.

To determine the holes of the function, we will first have to determine if there are any terms we can divide out from both the numerator and the denominator. To do this, we will factor the numerator and denominator. We begin with the numerator. First, note that each term is divisible by $$x$$, so we will factor out $$x$$. $(x^3-x^2-2x=x(x^2-x-2)$). Now we will factor $$x^2-x-2$$. We will do this by multiplying the leading coefficient by the constant term and determining its factors. $(1\cdot (-2)=-2$) The whole-number pairs of factors of $$-2$$ are $$\{1,-2\}$$ and $$\{-1,2\}$$. We want to know which pair of factors will add up to -1, the coefficient of the $$x$$ term. -2+1=-1, so we will use the pair of factors $$\{1,-2\}$$ to rewrite $$-x$$ as $$-2x+x$$ Then we will rewrite the quadratic expression as $(x^2-x-2=x^2-2x+x-2$) Now we will separate the quadratic expression by its first two terms and its last two terms to get $((x^2-2x)+(x-2)$) Next, we will factor out any common terms in each expression in parentheses, giving us $(x(x-2)+(x-2)$) This is equal to $( (x+1)(x-2)$) which is the factored form of the quadratic expression $$x^2-x-2$$. That means that the factored form of the numerator is $$x(x+1)(x-2)$$. Now we will factor the denominator, $$3x^2+x-14$$. Since it is already a quadratic expression, we will use the same method we used above for $$x^2-x-2$$. The product of the leading coefficient and the constant term is $$3\cdot(-14)=-42$$. We will now list the pairs of whole-number factors of -42: 1 and -42 2 and -21 3 and -14 6 and -7 -1 and 42 -2 and 21 -3 and 14 -6 and 7 We want to choose the pair that adds up to 1, which is the coefficient of the $$x$$ term, so we will pick -6 and 7 to rewrite $$x=-6x+7x$$ This gives us $(3x^2+x-14=3x^2-6x+7x-14=(3x^2-6x)+(7x-14)$) Now we will factor out common terms of the expressions in parentheses, giving us $((3x^2-6x)+(7x-14)=3x(x-2)+7(x+2)=(3x+7)(x-2)$) This means that the factored form of the denominator is $$(3x+7)(x-2)$$ Using this information, we can rewrite the function as $(f(x)=\frac{x(x-2)(x+1)}{(x-2)(3x+7)}$) Since $$x-2$$ is in the numerator and denominator, we know that there is a hole where $$x-2=0$$, so there is a hole at $$x=2$$. To find the $$x$$-intercepts, we need to know where $$y=0$$. $$y=0$$ when the numerator is equal to zero but the denominator is not. First, we will find where the numerator is equal to zero. Let $(x(x-2)(x+1)=0$) We will consider the case where each term is equal to zero. This tells us that the solutions are $(x=0$) $(x-2=0\text{, which means }x=2$) and $(x+1=0\text{, which means } x=-1$) This means that the numerator is equal to zero for $$x=0,2,-1$$. By plugging in each value to the denominator, we find that the denominator is equal to zero when $$x=2$$ (which makes sense, since this has been determined to be a hole), and that it is not equal to zero when $$x=0$$ and $$x=-1$$. This means that the $$x$$-intercepts are (0,0) and (-1,0). To find the $$y$$-intercepts, we will determine where $$x=0$$. $$f(0)=\frac{0(0-2)(0+1)}{(0-2)(3(0)+7)}=\frac{0}{-14}=0$$. Therefore, the only $$y$$-intercept is $$(0,0)$$. Now we will find the asymptotes of the function. First, we will look for vertical asymptotes. The function will have vertical asymptotes wherever the denominator is equal to zero and there is no hole. To determine where the denominator equals zero, we will solve the equation $$(x-2)(3x+7)=0$$. This will hold when $$x-2=0$$, which means $$x=2$$ and where $$3x+7=0$$, which means $$3x=-7$$, or, equivalently, $$x=-\frac{7}{3}$$ Since there is a hole at $$x=2$$, the only vertical asymptote is at $$x=-\frac{7}{3}$$. Since the degree of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. Finally, to find any other asymptotes, we will divide the numerator by the denominator and find the remainder using long division. This gives us $(\frac{x^3-x^2-2x}{3x^2+x-14}=\frac{1}{3}x-\frac{4}{9}+\frac{-\frac{28}{9}x-\frac{56}{9}}{3x^2+x-14}$) This means that there is a slant asymptote defined by $$y=\frac{1}{3}x-\frac{4}{9}$$. To summarize our findings, we have that there is a hole at $$x=2$$, the $$x$$-intercepts are (0,0) and (1,0), there is a $$y$$-intercept at (0,0), there are no horizontal asymptotes, there is a vertical asymptote at $$x=-\frac{7}{3}$$, and there is a slant asymptote at $$y=\frac{1}{3}x-\frac{4}{9}$$

### Subject:Calculus

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Question:

Find the derivative of $(f(x)=x^2+4$) using the limit definition of the derivative.

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Victoria B.

We recall that $(f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$) We will now plug in $$x^2-4$$ as $$f(x)$$, which gives us $(f'(x)=\lim_{h\to 0}\frac{(x+h)^2-4-(x^2-4)}{h}$) Distributing the negative sign, we get $(f'(x)=\lim_{h\to 0}\frac{(x+h)^2-4-x^2+4}{h}=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}$) Expanding the $$(x+h)^2$$ term gives us $(f'(x)=\lim_{h\to 0}\frac{(x^2+2xh+h^2)-x^2}{h}=\lim_{h\to 0}\frac{h^2+2xh}{h}$) Dividing the $$h$$ term, we get $( f'(x)=\lim_{h\to 0} (h^2+2x)$) Now we can apply the limit to get $(f'(x)=0^2+2x=2x$). This means that, when $$f(x)=x^2-4$$, $(f'(x)=2x$).

### Subject:Algebra

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Question:

What are the solutions to the equation $$12x^2+32x+12=-4$$?

Inactive
Victoria B.

We notice that this is a quadratic equation, so we will be able to apply the quadratic formula to find the solutions. To do this, we need to first get it in the form $(ax^2+bx+c=0$) To do this, we will add 4 to both sides. This gives us $(12x^2+32x+16=0$) which is in the desired form. Now, we notice that all terms are divisible by 4, so we will divide both sides by 4 to simplify the equation, giving us $(3x^2+8x+4=0$) This is still of the form $$ax^2+bx+c=0$$,where $$a=3$$, $$b=8$$, and $$c=4$$. Next, we will apply the quadratic formula. $(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$) Plugging in our values for $$a$$, $$b$$, and $$c$$ gives us $(x=\frac{-8\pm\sqrt{8^2-4(3)(4)}}{2(3)}$) Applying the exponent and performing multiplication gives us $(x=\frac{-8\pm \sqrt{64-48}}{6}$) Simplifying the term in the radical, we get $(x=\frac{-8\pm \sqrt{16}}{6}$) Finally, applying the square root gives us $(x=\frac{-8\pm4}{6}$) Now we will consider the two cases separately, since we have that $(x=\frac{-8+4}{6}\text{ or }x=\frac{-8-4}{6}$) In the first case, $(x=\frac{-8+4}{6}=\frac{-4}{6}=-\frac{2}{3}$) In the second case, $(x=\frac{-8-4}{6}=\frac{-12}{6}=-2$) This means that the solutions to the equation are $(\{-\frac{2}{3},-2\}$)

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