MATH 31 LESSONS Chapter 4: Max / Min Chapter 5: Sketching 3. Sketching Polynomials

Steps for Sketching Polynomial Functions Step 1. Degree State the degree of f (x) and identify the shape.

When the degree is even, then:  if a > 1, it opens up Deg = 2 Deg = 4

When the degree is even, then:  if a > 1, it opens up  if a < 1, it opens down Deg = 2 Deg = 4 Deg = 2 Deg = 4

When the degree is odd, then:  if a > 1, it rises to the right Deg = 3 Deg = 5

When the degree is odd, then:  if a > 1, it rises to the right  if a < 1, it falls to the right Deg = 3 Deg = 5

Step 2. Intercepts y-intercept(s) Let x = 0, and then solve for y x-intercept(s) Let y = 0, and then solve for x

Note: To find the x-intercept:  Factor completely and find the zeros - for complex functions, you may need to use the factor theorem and long division

To find the x-intercept:  For second degree factors of the form Ax 2 + Bx + C = 0, you can use the quadratic formula to solve. i.e.

To find the x-intercept:  If you can’t factor or use the quadratic formula, then use Newton’s method for finding roots. i.e. Take a first guess x 1. Then, to find x 2 :

Step 3. First Derivative Test  Differentiate and state the critical values i.e. When f (x) = 0

 Differentiate and state the critical values i.e. When f (x) = 0  Use the interval test to show where the function is increasing and decreasing

 Differentiate and state the critical values i.e. When f (x) = 0  Use the interval test to show where the function is increasing and decreasing  Identify local (and absolute) max / mins - substitute into the original function to get the y-values

Step 4. Sketch the Function  Place all intercepts and critical values on the grid  Using your knowledge of where the function is increasing and decreasing, connect the dots  Extend the arms on either side to infinity - recall that polynomial functions are continuous and have a domain x  

Ex. 1Sketch the following function: Try this example on your own first. Then, check out the solution.

 Degree: This function is of degree 3 and a > 0. Thus, this function will rise to the right.

 Intercepts: y-intercept: (x = 0) So, there is a y-intercept at (0, 0).

x-intercepts: (y = 0) So, there are x-intercepts at (0, 0) and (  6.93, 0)

 First Derivative Test:

Find CV’s

Interval test: 4 f -4 Sketch a number line, using the CV’s as boundaries

x < -4: Since f > 0, it is increasing. e.g. x = -5 4 f -4 (  ) (  )

-4 < x < 4: Since f < 0, it is increasing. e.g. x = 0 4 f -4 (  ) (  ) (+) (  )

x > 4: Since f > 0, it is increasing. e.g. x = 5 4 f -4 (  ) (  ) (+) (  ) (+) (+)

Find the y-values by subbing them into the original function: Local Min Local Max 4 f -4 (  ) (  ) (+) (  ) (+) (+)

 Sketch: y x First, put the intercepts and the CV’s on the graph -44

y Then, use the interval test to connect the dots. 4 f x -44

Ex. 2Sketch the following function: Try this example on your own first. Then, check out the solution.

 Degree: This function is of degree 4 and a > 0. Thus, this function will open up.

 Intercepts: y-intercept: (x = 0) So, there is a y-intercept at (0, 4).

x-intercepts: (y = 0) So, there are x-intercepts at (  1, 0) and (  2, 0)

 First Derivative Test:

Interval test: Sketch a number line, using the CV’s as boundaries 0 f

x < -1.58: Since f < 0, it is decreasing. e.g. x = -3 0 f (  ) (+) 1.58

-1.58 < x < 0: Since f > 0, it is increasing. e.g. x = -1 0 f (  ) (+) (  ) (  ) 1.58

0 < x < 1.58: Since f < 0, it is decreasing. e.g. x = 1 0 f (  ) (+) (  ) (  ) (+) (  ) 1.58

x > 1.58: Since f > 0, it is increasing. e.g. x = 3 0 f (  ) (+) (  ) (  ) (+) (+) (+) (  ) 1.58

Find the y-values by subbing them into the original function: Local Min Local Max 0 f (  ) (+) (  ) (  ) (+) (+) (+) (  ) 1.58 Local Min

 Sketch: y x First, put the intercepts and the CV’s on the graph (-1.58, -2.25)(1.58, 2.25)

Then, use the interval test to connect the dots. 0 f y x (-1.58, -2.25)(1.58, 2.25)