# Curve Sketching of Polynomials

Curve sketching requires you to extract all the knowledge about the following points:

1. Stationary points
2. Points of inflexion
3. x- and y- intercepts
4. Domain and Range
5. Asymptotes or Limits
6. Symmetry, Odd or Even Functions
7. Table of Values

These prerequisite knowledge are applied (not necessarily to all) in curve sketching. Sure, quadratics are not hard to sketch, but polynomials need more calculations to obtain sufficient information for sketches.

Let's start with one example: $f(x)={x}^{3}-3{x}^{2}-9x+2$.

First, we could find the stationary points and points of inflexion.

$f'(x)=3{x}^{2}-6x-9$  and  $f''(x)=6x-6$

First, find the stationary points:

$f'(x)=0 \\ 3{x}^{2}-6x-9=0 \\ {x}^{2}-2x-3=0 \\ (x-3)(x+1)=0 \\ \therefore \quad x=-1 or 3 \\ \\ f(3)={(3)}^{3}-3{(3)}^{2}-9(3)+2 \\ f(3)=-25$

So (3, -25) is a stationary point.

$f(-1)={(-1)}^{3}-3{(-1)}^{2}-9(-1)+2$

f(-1)=7 \\ So (-1, 7) is also a stationary point.

Then, the second derivative determine their nature of stationary points.

$f''(3)=6(3)-6=12 > 0$ (concave upwards)

$\therefore \quad (3, -25)$ is a minimum turning point.

$f''(-1)=6(-1)-6=-12 < 0$ (concave upwards)

$\therefore \quad (-1, 7)$ is a maximum turning point.

Are there any points of inflexion? Let's check it out with the second derivative = 0.

$f''(x)=0 \\ 6x-6=0 \\ 6x=6 \\ x=1$

$f(1)={(1)}^{3}-3{(1)}^{2}-9(1)+2=-9$

Check for change of concavity:

When x=0, f''(0)=-6

When x=1, f''(1)=0

When x=2, f''(2)=6

Since concavity changes, (1, -9) is a point of inflexion.

Next we check the intercept:

For y-intercept, x=0:

$f(0)={(0)}^{3}-3{(0)}^{2}-9(0)+2=2$

Therefore, the y-intercept is on (0, 2).

The x-intercepts are hard to solve, however we have enough information to sketch this curve.

We understand that the maximum  and minimum points are not THE maximum or minimum point of the whole function, therefore we call the points relative maximum and relative minimum.

### Examples

Sketch the curve $y={(x+1)}^{2}(x-3)$.

First we can find the intercepts of the curve:

When y=0,

${(x+1)}^{2}(x-3)=0$

So, either

${(x+1)}^{2}=0 \\ x+1=0 \\ x=-1$

Or

$x-3=0 \\ x=3$

When x=0,

$y={(0+1)}^{2}(0-3)=0$

Therefore, the x-intercepts are (-1, 0) and (3, 0), and the y-intercept is (0, -3).

Now, to check their relative maximum and relative minimum:

$y=({ x }^{ 2 }+2x+1)(x-3)\\ y={ x }^{ 3 }+2{ x }^{ 2 }+x-3{ x }^{ 2 }-6x-3\\ y={ x }^{ 3 }-{ x }^{ 2 }-5x-3\\ \\ y'=3{ x }^{ 2 }-2x-5\\ \\ y''=6x-2$

Check the stationary points:

$y'=3{ x }^{ 2 }-2x-5=0$

${ x }_{ 1,2 }=\cfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ { x }_{ 1,2 }=\cfrac { -(-2)\pm \sqrt { (-2)^{ 2 }-4(3)(-5) } }{ 2(3) } \\ \\ { x }_{ 1,2 }=\cfrac { 2\pm \sqrt { 64 } }{ 6 } \\ \\ { x }_{ 1 }=\cfrac { 2+8 }{ 6 } \qquad \qquad \qquad { x }_{ 2 }=\cfrac { 2-8 }{ 6 } \\ \\ { x }_{ 1 }=\cfrac { 10 }{ 6 } =\cfrac { 5 }{ 3 } \qquad \qquad { x }_{ 2 }=\cfrac { -6 }{ 6 } =-1$

When x=$\cfrac{5}{3}$y=$-9\cfrac{14}{27}$

When x=-1, y=0 (this stationary point touches the x-axis)

Therefore, the stationary points are: $(\cfrac{5}{3}$$-9\cfrac{14}{27})$ and (-1, 0).

Check for inflexion point:

$y''=6x-2=0 \\ 6x-2=0 \\ 6x=2 \\ x=\cfrac{2}{6} \\ x=\cfrac{1}{3}$

When x=$\cfrac{1}{3}$y=$-\cfrac{32}{27}$

Therefore, the inflexion point is: $(\cfrac{1}{3}$$-\cfrac{32}{27})$.

The component ${(x+1)}^{2}$, which was supposed to be 2 roots became one root that touches the x-axis. The power of 2 gives us an indication that the curve touches the x-axis.

But what happens if the function was instead: $y={(x+1)}^{3}(x-3)$. The power of 3 on ${(x+1)}^{3}$ creates a point of inflexion, and in this case, there are two points of inflexion. Have a look at the diagram below.

Try the curve sketching above with proper calculations.

Sketch the curve $y=\cfrac{x+2}{{x}^{2}-4}$.

Simplifying the equation:

$y=\cfrac { x+2 }{ (x+2)(x-2) }$ difference of two squares)

$y=\cfrac { 1 }{ x-2 }$

Check the domain and range:

Domain:

$x-2 \neq 0 \\ x \neq 2$

The vertical asymptote is therefore: $x=2$

Range (with limits):

$x\rightarrow { 0 }^{ + },\quad y\rightarrow \infty \\ x\rightarrow { 0 }^{ - },\quad y\rightarrow \infty \\ x\rightarrow \infty ^{ + },\quad y\rightarrow 0\\ x\rightarrow \infty ^{ - },\quad y\rightarrow 0$

The horizontal asymptote here is: $y=0$

We can find the intercepts:

When x=0, y= $-\cfrac{1}{2} \rightarrow (0, -\cfrac{1}{2})$

There are no x-intercepts.

Sketch the curve $f(x)=\cfrac{{x}^{2}}{{x}^{2}-4}$.

A quick way to sketch this curve is by analysing its asymptotes, intercepts and sections.

Find the asymptotes:

Vertical asymptote:

${x}^{2}-4 \neq 0 \\ {x}^{2} \neq 0 \\ {x}^{2}=4$

Therefore, the vertical asymptote is $x=\pm 2$.

Horizontal asymptote:

To find the horizontal asymptote we can use limits, and in this case there is another approach that we can use - dividing all numbers by ${x}^{2}$

$y=\lim _{ x\rightarrow \infty }{ \cfrac { { x }^{ 2 } }{ { x }^{ 2 }-4 } } \\ \\ y=\lim _{ x\rightarrow \infty }{ \cfrac { \cfrac { { x }^{ 2 } }{ { x }^{ 2 } } }{ \cfrac { { x }^{ 2 } }{ { x }^{ 2 } } -\cfrac { 4 }{ { x }^{ 2 } } } } \\ \\ y=\lim _{ x\rightarrow \infty }{ \cfrac { 1 }{ 1-\cfrac { 4 }{ { x }^{ 2 } } } } \\ \\ y=\cfrac { 1 }{ 1-0 } \\ \\ y=1$

Therefore, as x goes to $\infty$ y approaches 1.

Intercepts are easier to find here:

When x=0, y=0

When y=0, x=0 (obviously from the previous)

So far, we understand that the asymptotes and intercepts give some indication of how to draw the curve:

To find the Sections of the curve, we check on the values using table of values.

We can see the symmetrical pattern of the curve and now the information is sufficient to sketch the curve.