Transcript File

Tangents and Gradients
Points with a Given Gradient
e.g. Find the coordinates of the points on the curve
y  x 3  8 x  7 where the gradient equals 4
Gradient of curve
= gradient of tangent
= 4
We need to be able
to find these points
using algebra
Points with a Given Gradient
e.g. Find the coordinates of the points on the curve
y  x 3  8 x  7 where the gradient is 4
Solution:
The gradient of the curve is given by
y  x  8x  7 
3
dy
4
dx
Gradient is 4 

3x  8  4
2
dy
 3x2  8
dx
Quadratic equation
with no linear x -term

3 x 2  12
 x  4  x  2
3
x  2  y  ( 2)  8( 2)  7  1
3
x   2  y  ( 2)  8( 2)  7  15
2
dy
dx
Points with a Given Gradient
The points on
y  x  8 x  7 with gradient 4
3
(2, 15) x
x
( 2,  1)
SUMMARY
 To find the point(s) on a curve y  f ( x ) with a
given gradient:
•
find the gradient function dy
dx
•
dy
let
equal the given gradient
dx
•
solve the resulting equation
Exercises
Find the coordinates of the points on the curves with
the gradients given
1.
y  x2  4x  3
where the gradient is -2
Ans: (-3, -6)
3
2
2. y  x  3 x  21 x  20 where the gradient is 3
( Watch out for the common factor in the
quadratic equation )
Ans: (-2, 2) and (4, -88)
Increasing and Decreasing Functions
•
An increasing function is one whose
gradient is always greater than or equal to
zero.
dy
 0
dx
•
for all values of x
A decreasing function has a gradient that
is always negative or zero.
dy
 0
dx
for all values of x
e.g.1 Show that y  x 3  4 x is an increasing
function
dy
3
 3x2  4
Solution: y  x  4 x 
dx
dy is the sum of
dx • a positive number ( 3 )  a perfect
•
square ( which is positive or zero for
all values of x, and
a positive number ( 4 )
dy

 0 for all values of x
dx
so, y  x 3  4 x is an increasing function
e.g.2 Show that y  1 x 3  3 x 2  9 x is an
3
increasing function.
Solution: y 
1
3
x  3x  9x 
3
2
dy
 x2  6x  9
dx
To show that x  6 x  9 is never negative ( in
spite of the negative term ), we need to
complete the square.
2
x 2  6 x  9  ( x  3) 2
Since a square is always greater than or equal
to zero,
dy
 0 for all values of x
dx
2
1 3
y

x

3
x
 9 x is an increasing function.

3
The graphs of the increasing functions
y  x 3  4 x and y  13 x 3  3 x 2  9 x
y  x3  4x
and
y  13 x 3  3 x 2  9 x
are
Exercises
1. Show that y   x 3 is a decreasing function and
sketch its graph.
2. Show that y  1 x 3  2 x 2  5 x is an increasing
3
function and sketch its graph.
Solutions are on the next 2 slides.
Solutions
1. Show that y   x 3 is a decreasing function and
sketch its graph.
dy
Solution:
 3x 2 . This is the product of a
dx
square which is always  0 and a negative number,
dy
so
 0 for all x. Hence y   x 3 is a
dx
decreasing function.
y  x3
Solutions
2. Show that y  1 x 3  2 x 2  5 x is an increasing
3
function and sketch its graph.
dy
Solution:
 x2  4x  5 .
dx
dy
Completing the square:
 ( x  2) 2  1
dx
which is the sum of a square which is  0
and a positive number. Hence y is an increasing
function.
y  13 x 3  2 x 2  5 x
The equation of a tangent
e.g. 1 Find the equation of the tangent at the point
(-1, 3) on the curve with equation
y  x3  3x2  2x  1
3
2
Solution:
y

x

3
x
 2 at
x a1 point and the gradient of
The gradient of a curve
dy the tangent
Gradient
=equal
-5
2
at
that
point
are

 3x  6x  2
dx
 At x = 1
(-1, 3) x
m  3( 1) 2  6( 1)  2
 5
y  mx  c  y  5 x  c
(-1, 3) on line:  3  5( 1)  c   2  c
So, the equation of the tangent is y  5 x  2
An Alternative Notation
The notation f ( x ) for a function of x can be used
instead of y.
dy
When f ( x ) is used, instead of using
for the
dx
gradient function, we write
f / ( x ) ( We read this as “ f dashed x ” )
This notation is helpful if we need to substitute for x.
e.g.
f ( x)  x 3  2 x 2  3 x  1

f / ( x)  3 x 2  4 x  3

f / ( 2 )  3( 2) 2  4( 2)  3  17
e.g. 2 Find the equation of the tangent where x = 2 on
the curve with equation y  f ( x ) where
f ( x)  x  2 x  3 x  4
3
2
Solution: To use y  mx  c we need to know y at
the point as well as x and m
f ( x )  x  2 x  3 x  4      (1)
y  f ( 2)  ( 2) 3  2( 2) 2  3( 2)  4
 8864  2
From (1), f / ( x )  3 x 2  4 x  3
m  f / ( 2)  3( 2) 2  4( 2)  3
 12  8  3  7
y  mx  c 
y  7x  c
(2, 2) on the line  2  7( 2)  c   12  c
So, the equation of the tangent is y  7 x  12
3
2
SUMMARY
 To find the equation of the tangent at a point on
the curve y  f ( x ) :
• if the y-value at the point is not given,
substitute the x -value into the equation of the
curve to find y
•
•
•
dy
find the gradient function
( or f / ( x ) )
dx
substitute the x-value into dy ( or f / ( x ) )
dx
to find the gradient of the tangent, m
substitute for y, m and x into
y  mx  cto find c
Exercises
1.
Find the equation of the tangent to the curve
y  x 3  2 x 2  x  3 at the point (2, -1)
Ans: y  5 x  11
2.
Find the equation of the tangent to the curve
y  f ( x ) at the point x = -1, where
f ( x )  x 3  3 x 2  11 x  6
Ans: y  2 x  1
The equation of a tangent
e.g. 1 Find the equation of the tangent at the point
(-1, 3) on the curve with equation
y  x3  3x2  2x  1
Solution: y  x 3  3 x 2  2 x  1
Gradient = -5
dy
2

 3x  6x  2
dx
 At x = -1
(-1, 3) x
m  3( 1) 2  6( 1)  2
 5
y  mx  c  y  5 x  c
(-1, 3) on line:  3  5( 1)  c   2  c
So, the equation of the tangent is y  5 x  2
e.g. 2 Find the equation of the tangent where x = 2 on
the curve with equation y  f ( x ) where
f ( x)  x  2 x  3 x  4
Solution: To use y  mx  c we need to know y at
3
2
the point as well as x and m
f ( x )  x  2 x  3 x  4      (1)
y  f ( 2)  ( 2) 3  2( 2) 2  3( 2)  4
 8864  2
From (1), f / ( x )  3 x 2  4 x  3
m  f / ( 2)  3( 2) 2  4( 2)  3
 12  8  3  7
y  mx  c 
y  7x  c
(2, 2) on the line  2  7( 2)  c   12  c
So, the equation of the tangent is y  7 x  12
3
2