Graphs Lesson 2 Aims: • To be able to use graphical calc to investigate graphs of rational functions • To be able to draw graphs of rational functions of the forms: and • To be able to use the discriminant properties • To be able to find stationary points on these sketched graphs

Starter What is the discriminant? We will come back to use these ideas in the second half of the lesson What is the discriminant? What does it tell us? How many roots has x2 + x – 6 = 0 ? Question Find the possible of k such that the quadratic has 2 equal roots. kx2 + 3x + 2 = 0

1. Find the intercepts with the axes Recap Summary of Steps 1. Find the intercepts with the axes 2. Find the vertical asymptotes 3. Examine the behaviour as x tends to   Properties of today’s types of graphs! ; only 1 vertical and 1 horizontal asymptote. ; will have no vertical asymptote and 1 horizontal. Denominator is a r_____________ factor Often this graph cuts the horizontal asymptote Where the denominator is an i________________ quadratic

1. Find the intercepts with the axes Sketch If x = 0, y = y = 0, x = 2. Find the vertical asymptotes (x – 2)2 = 0 when x = this is the only vertical asymptote 3. Find the Examine the behaviour as x tends to   x    y  

So y = 1 is a horizontal asymptote Notice the graph has to cut the horizontal asymptote. This happens at the point where y = and x =

1. Find the intercepts with the axes Sketch If x = 0, y = y = 0, ( )( ) = 0, so x = 2. Find the vertical asymptotes x² + 2x + 6 = 0 has no real solutions so there are no vertical asymptotes 3. Examine the behaviour as x tends to   x    y  

So y = 1 is a horizontal asymptote

On w/b Do Exercise 5D page 63 And Exercise 5E page 64 2. Sketch

On w/b Do Exercise 5D page 63 And Exercise 5E page 64 2. Sketch

Finding a Stationary Point Consider the line y = k. This is a horizontal line. When it move up to touch the stationary point, how many times will it cross the curve? _______ What is another name for the line at this point?__________ To find where they crossed, we would equal the line to the curve and create an impossible equation to solve, as it has two unknowns, x and k. However, what do we know about the value of the discriminant here?________________ Now we can find k and go on to find where they cross. y = k

Finding a Stationary Point y = k

Finding a Stationary Point * Sub this value of k into * and then use your graphical calculator to solve the polynomial for the x value:

Maximum Stationary point is ( ) w/b qu. together first Do exercise 5F page 68 Questions 2-3 no need to sketch the graph. 4-5 need sketches!

On w/b Find the stationary point on by equating it with the line y = k For tangent b2 – 4ac = 0

On w/b 1. Find the stationary point on For tangent b2 – 4ac = 0

Do exercise 5F page 68 Questions 2-3 no need to sketch the graph. 4-5 need sketches!