# Tutor profile: Narayan N.

## Questions

### Subject: Pre-Calculus

Consider the function $$f(x)=x^2+1$$, find the difference quotient, and explain what we get from the difference quotient.

Here, $$f(x)=x^2+1$$ We need to find the difference quotient. The formula for the difference quotient is $$\frac{f(x+h)-f(x)}{h}$$ Now, $$f(x+h)=(x+h)^2+1$$ $$=(x+h)(x+h)+1$$. [Because $$(x+h)^2=(x+h)(x+h)$$] $$=x^2+2xh+h^2+1$$ [Using FIOL $$(x+h)^2=x^2+2xh+h^2$$] We will now plug it into the difference quotient formula $$\frac{f(x+h)-f(x)}{h}$$ =$$\frac{x^2+2xh+h^2+1-(x^2+1)}{h}$$ $$=\frac{x^2+2xh+h^2+1-x^2-1}{h}$$ $$=\frac{2xh+h^2}{h}$$. [ Simplifying the Numerator] $$=\frac{2xh}{h}$$+$$\frac{h^2}{h}$$. [Separating the Denominator] $$2x+h$$ Hence, the difference quotient is $$\frac{f(x+h)-f(x)}{h}$$=$$2x+h$$ Now, we need to interpret the meaning of the difference quotient. The difference quotient gives the average rate of change on the interval $$[x,x+h]$$. For example, if a particle is moving along the path $$y=f(x)$$, then the difference quotient will give the average velocity of the particle on the interval $$[x,x+h]$$. If the value of h approaches zero or the interval is very small, the difference quotient will give the instantaneous rate of change (or the instantaneous velocity of the particle moving on the curve $$y=f(x)$$).

### Subject: Trigonometry

In a circle of radius $$150cm$$, what is the length of an arc subtended by an arc a central angle of $$45^o$$?

Solution Here, the radius of the circle $$(r)= 150 cm$$ central angle $$\theta$$=$$45^o$$ Now, the formula to find the arc length is $$s$$= $$r\times\theta$$ where $$\theta$$ is in radian Here central angle is $$45^o$$, we need to change it to radian $$\theta$$=45 ($$ \frac{\pi}{180} $$ )=$$ \frac{\pi}{4} $$ Now, sunstuting the values in to the formula for arc length $$s=$$ 150$$\times$$ $$\frac{\pi}{4}$$ =$$37.5\pi$$ =171.81 Hence, the length of the arc is 171.81 cm

### Subject: Algebra

Suppose $$f(x)=2x^2-4x+3$$ is the quadratic function given. Write the equation in vertex form and state its vertex and the equation of the line of symmetry.

Solution Here , $$f(x)=2x^2-4x+3$$ and we want to write it as $$f(x)=a(x-h)^2+k$$ Factor out 2 from the first two terms, $$f(x)=2(x^2+2x)+3$$ Now, in order to complete the square for $$x^2+2x$$ we add and subtract 1 So, $$f(x)=2(x^2-2x+1-1)+3$$ $$f(x)=2(x^2-2x+1)-2+3$$. [Simplifying by Distributing 2] $$f(x)=2(x^2-2x+1)+1$$. [Simplifying 3-2=1} $$f(x)=2(x-1)^2+1$$ [ Writing $$x^2-2x+1$$ as $$(x-1)^2$$] Hence, the vertex form is $$f(x)=2(x-1)^2+1$$ with $$h=1$$ and $$k=1$$ So, the vertex $$(h,k)=(1,1)$$ and the equation of the line of symmetry is $$x-1=0$$ or $$x=1$$