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# Tutor profile: Luis T.

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Luis T.
Math Tutor for 5 Years
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## Questions

### Subject:Pre-Calculus

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Question:

If I give the following parabola y=2x^2-2x+1, find the vertex of this equation.

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Luis T.

To find the vertex of a parabola you need two equations. Suppose parabola equation is y=ax^2+bx+c , To find the vertex it has the form (-b/2a,f(-b/2a)). To find these values you just plug in. -b/2a=2/2=1, f(-b/2a) is we will plug in 1 in every x you see so you get f(1)=2-2+1=1. Therefore, your vertex is (1,1).

### Subject:Trigonometry

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Question:

Suppose that we have a right triangle with adjacent side being 4in and opposite side to be 3in. Find the angle above adjacent side.

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Luis T.

There are three formulas you can use to find specific angles based on what sides you have. cos(theta)=adjacent/hypotenuse , sin(theta)=opposite/hypotenuse , or tan(theta)=opposite/adjacent. Since both opposite and adjacent sides are given you use tan(theta). Therefore tan(theta)=3/4, to solve for theta you take tan inverse of each side and you get theta=taninverse(3/4) and plug into calculator.

### Subject:Calculus

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Question:

Take the derivative of f(x)=x^3+2x^2-3x

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Luis T.

This type of function is a x^n function. To take the derivative of x^n is as follows: d/dx[x^n] = nx^(n-1). Therefore d/dx[x^3] = 3x^2, d/dx[2x^2]=4x, d/dx[-3x]=-3. So, f'(x)=3x^2+4x-3

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