# How do I sketch the graph y = (x^2 + 4*x + 2)/(3*x + 1)

This is an example of a rational function. So to sketch it, we need to know three things:

1) where it crosses the x and y axis

2) Where its turning points are, if it has any

3) Where its asymptotes are, if it has any.

Let's start with 1):

The graph crosses the y axis when x=0. So if x=0, then y=4/1 = 4. So the graph crosses the y axis at (0,4)

The graph crosses the x axis when y=0, which means the numerator of the fraction (x^{2} + 4*x + 4)/(3*x + 2) = 0. We see that the numerator can be factorised into (x + 2)^{2}, which means that the numberator only equals zero when x = -2. So the graph crosses the x axis at (-2,0)

Now onto 2):

Using the quotient rule, dy/dx = (3*x^{2} + 4*x - 4)/(3*x + 2)^{2}

Now this is zero when the numerator is zero. We can factorise the numberator into (x+2)*(3*x - 2) to see that dy/dx is zero when x = -2 or x = 2/3. We could have also used the quadratic formula to work this out.

When x = -2, y = 0 as we already established. So this is where we cross the x axis and a turning point at the same time.

When x = 2/3, y = (2/3 + 2)^{2}/(3*(2/3) + 1) = 64/27

by differentiating again to find dy/dx we can classify each stationary point: (-2,0) is a maximum and (2/3,64/27) is a minimum.

Now for part 3)

An asymptote is when the graph "shoots off to inifinity". So it occurs when the denominator of the function is zero. Here that only occurs when (3*x+1) = 0, or when x = -1/3.

So we have the parts of our graph we need to draw it. The curve comes in from the bottom right, touches the x axis at (-2,0) to change direction, shoots off to negative infinity along the line x=-1/3 (but never crossing it). Emerges from positive infinity the other side of the line x=-1/3, turns at the point (2/3,64/27) and swoops off the top right corner.