Further Pure 1 Lesson 7 – Sketching Graphs

Wiltshire Graphs and Inequalities Good diagrams Communicate ideas efficiently. Helps discovery & understanding of relationships. Essential Features include Points where axes are crossed. Nature and position of turning points. Behaviour as x or y tend to infinity. Need to be able to check using a calculator or computer but must be able to do it independently.

Wiltshire Rational Functions A rational function is defined as a function N(x)/D(x), where N(x) & D(x) are polynomials. Note that D(x) = 0 (This is similar to the notion of a rational number, n/d) What does the graph of y = 1/x look like? What happens when x = 0? Now what would the graph of y = 1/(x-2)+3 look like.

Wiltshire Using Transformations Now what would the graph of y = 1/(x-2)+3 look like. This will look the same as y = 1/x but after two transformations. The (x – 2) shifts the graph right 2. Then the +3 shifts the graph up 3. Notice where the asymptotes have moved too. This would be the same as y = (3x – 5)/(x - 2)

Wiltshire Using Transformations How could we plot the graph y = (2x+9)/(x+4) y = 1/(x+4)+2 Or 2 x + 4 2x + 9 2x We can now use transformations to plot the graph.

Wiltshire Questions Plot the following graphs i) ii) iii)

Wiltshire Vertical Asymptotes An asymptote is a straight line which a curve approaches tangentially as x or y tends to infinity. Asymptotes are usually shown on a graph by a dotted line. In the graph y = 1/x there is an asymptote at x = 0. This is because as x gets smaller and smaller y tends to infinity. (Note: it can help that 1/0 is undefined) What would the vertical asymptote be for the graph y = 1/(x – 3)? x = 3. In general the asymptote for y = 1(x - a) will be x = a. and the asymptote for y = 1/(x + a) will be x = -a.

Wiltshire Vertical Asymptotes What are the vertical asymptotes for the following graphs? i) ii) iii)

Wiltshire Sketching Graphs We are now going to look at the steps for sketching the graph. There are some steps to follow to ensure that the graph is plotted correctly. 1)Find where the graph cuts the axes. 2)Find the vertical asymptotes and examine the behaviour near them. 3)Examine the behaviour as x tends to infinity. 4)Complete the sketch

Wiltshire Sketching Graphs Step 1 - Find where the graph cuts the axes. The graph will cut the y-axis when x = 0, and the x- axis when y = 0. So when x = 0 And when y = 0 The graph will cut the axes at the co-ordinates (0,2) and (6,0).

Wiltshire Sketching Graphs The graph will cut the axes at the co-ordinates (0,2) and (6,0). We can plot these two co-ordinates on the graph.

Wiltshire Sketching Graphs Step 2 - Find the vertical asymptotes and examine the behaviour near them. The vertical limits will be when x = -3 & x = 1

Wiltshire Sketching Graphs There are 2 vertical asymptotes that can be approached from 2 directions. We need to look at each of the 4 scenarios.

Wiltshire Sketching Graphs i)When x is slightly less than -3. (x-6) is negative, (x+3) is negative and very small and (x-1) is negative.

Wiltshire Sketching Graphs ii)When x is slightly greater than -3. (x-6) is negative, (x+3) is positive and very small and (x-1) is negative.

Wiltshire Sketching Graphs i)When x is slightly less than 1. (x-6) is negative, (x+3) is positive and (x-1) is a very small negative.

Wiltshire Sketching Graphs i)When x is slightly greater than 1. (x-6) is negative, (x+3) is positive (x-1) is very small positive.

Wiltshire Sketching Graphs Step 3 - Examine the behaviour as x tends to infinity. As x tends to infinity y tends to zero from above the x-axis. As x tends to negative infinity y tends to zero from below the x-axis. So y = 0 is a horizontal asymptote.

Wiltshire Sketching Graphs Step 4 – Sketch the graph.

Wiltshire Sketching Graphs Notice that the graph is a smooth curve. There is a local minimum between x = -3 & 1. There is a local maximum just after x = 6. How do we know that there are only two turning points? We can find the exact position of the maxima and minima by solving dy/dx = 0. However you will not study this until C3.

Wiltshire Sketching Graphs An alternative way to look at it would be to consider the graph y equal to horizontal lines c. If y = (x-6)/((x+3)(x-1)) = c Then (x-6)=c(x+3)(x-1) Unless c = 0 then this is a quadratic equation with 3 possible outcomes. i)No solution ii)One solution iii)Two solutions As the quadratic cannot have more than 2 solutions, then the green line is impossible. This proves that our original drawing was correct. Can you explain how this would also show that the local maximum must be lower than the local minimum?

Wiltshire Sketching Graphs Sketch the following graphs? i) ii) iii)

Wiltshire Complicated example Sketch the following graph.

Wiltshire Using Symmetry Efficient knowledge of symmetrical curves can help you when sketching graphs. If f(x) = f(-x) then f(x) is symmetrical about the y-axis. Graphs of this type are known as even functions Note if f(x) = -f(-x) then f(x) has rotational symmetry of order 2 about the origin. Graphs of this type are known as odd functions. Name an even and an odd function.

Wiltshire Using Symmetry Sketch the graph Step 1 – When x = 0, y = 5 When y = 0 there is no x Step 2 – There is no x that makes x 2 +3 = 0 therefore no vertical asymptotes exist. Step 3 – As x tends to +ve/-ve infinity y tends to 2. So y = 2 is a horizontal asymptote. Step 4 – As x only contains even powers of x, y is an even function, and so is symmetrical about the y-axis.

Wiltshire Using Symmetry

Wiltshire Using a Graphics Calculator