Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

5

Finding the gradient from first principles

! Figure 5.6 shows Q in a position where h is positive, but negative values of

h would put Q to the left of P.

From figure 5.6, the gradient of PQ is ()^39 –

+h^2

h

= ++

= +

= +

=+

96 9

6

6

6

2

2

hh

h

hh

h

hh

h

h

–

()

.

For example, when h = 0.001, the gradient of PQ is 6.001, and when h = −0.001,

the gradient of PQ is 5.999. The gradient of the tangent at P is between these two

values. Similarly the gradient of the tangent would be between 6 − h and 6 + h for

all small non-zero values of h.

For this to be true the gradient of the tangent at (3, 9) must be exactly 6.

ACTIVITY 5.3 Using a similar method, find the gradient of the tangent to the curve at

(i) (1, 1)

(ii) (−2, 4)

(iii) (4, 16).

What do you notice?

The gradient function

The work so far has involved finding the gradient of the curve y = x^2 at a

particular point (3, 9), but this is not the way in which you would normally find

the gradient at a point. Rather you would consider the general point, (x, y), and

then substitute the value(s) of x (and/or y) corresponding to the point of interest.

h

Q

(3, 9) P

(3 + h)^2 – 9

(3 + h, (3 + h)^2 )

Figure 5.6