What have we done so far

1
What have we done so far?

• We have made a shape
• We have changed its size
• We have tried to find the optimum size
• using computer-generated tables
• using graphs
• using mathematical models (algebra) and
  mathematical graphs

2
What do we need to do now?

• We need to find an efficient way of finding the
  turning point(s) of any graph.

3
Activity three

• An investigation into graphs and their tangents

4
We ended the open box problem with a graph
This graph was generated using software called
Autograph
5
Using Autograph to draw the open box graph

• Type in the equation y x(24 2x)(20 2x)
• Change the scale so that
• - 10 lt x lt 20
• -500 lt y lt 1500
• Make the graph line thicker
• Put a point on to the graph
• Draw a tangent at the point
• Move the point around
• . . . and see what happens . . .

6
Notes

• Every point on the curve has a tangent
• A tangent is a straight line
• The tangent has its own equation
• The tangent has equation y mx c
• This equation is different for every position of
  the tangent

7
y 140x 576
y 256x 288
8
for tangent y 140x 576

• the gradient is 140
• the y-intercept is 576
• the point of contact of the tangent and the curve
  is (3 996)

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at the turning point . . .

• The tangent will be horizontal
• The gradient of the tangent will be ???
• 0

10
Take another example . . .

• Find the turning point(s) of the graph of
• y 2x3 - 3x2 - 12x 6

Use Autograph
11
focus on the gradient of the tangent
Gradient gt 0
Gradient lt 0
Gradient 0
Gradient 0
12
Calculus gives you a formula for the gradient
of the tangent
13
Graph formulay 2x3 3x2 12x 6

• Gradient formula
• 6x2 6x 12

Since the gradient of the tangent at the turning
point is 0 6x2 6x 12 0 x2 x 2 0 (x
2)(x 1) 0 x 2 or x 1
When x 2, y 2(2)3 3(2)2 12(2) 6
14 When x 1, y 2(1)3 3( 1)2 12( 1)
6 13
14
This means that we have calculated the
coordinates of the turning points of the graph
! ! ! ! USING CALCULUS ! ! ! !
(1 13)
(2 14)