Tutor profile: Luis T.
If I give the following parabola y=2x^2-2x+1, find the vertex of this equation.
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).
Suppose that we have a right triangle with adjacent side being 4in and opposite side to be 3in. Find the angle above adjacent side.
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.
Take the derivative of f(x)=x^3+2x^2-3x
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
needs and Luis will reply soon.