1. A(a,b) EQUATIONS OF TANGENTS y = mx +c y = f(x) tangent
NB: at A(a, b) gradient of line = gradient of curve
gradient of line = m (from y = mx + c )
gradient of curve at (a, b) = f (a)
it follows that m = f (a)

2. Example 24
Find the equation of the tangent to the curve y = x3 - 2x at the point where x = -1.
NB: tangent is line so we need point + gradient then we can use the formula y - b = m(x - a) .
Point: if x = -1 then y = (-1)3 - (2 X -1) + 1
= (-2) + 1
= 2 point is (-1,2)
Gradient: dy/dx = 3x2 - 2
when x = -1 dy/dx = 3 X (-1)2 - 2
= = 1
gradient = 1

3. Now using y - b = m(x - a)
we get y - 2 = 1( x + 1)
or y - 2 = x + 1
or y = x + 3

4. Example 25
Find the equation of the tangent to the curve y = x2 at the point where x = (x  0)
Also find where the tangent cuts the X-axis and Y-axis.
Point: when x = then y = (-2)2
= 4/4 = 1
point is (-2, 1)
Gradient: y = 4x so dy/dx = -8x-3
= x3
when x = -2 then dy/dx = (-2)3
= -8/-8 = 1
gradient = 1

5. Now using y - b = m(x - a)
we get y - 1 = 1( x + 2)
or y - 1 = x + 2
or y = x + 3
Axes
Tangent cuts Y-axis when x = 0
so y = = 3
at point (0, 3)
Tangent cuts X-axis when y = 0
so 0 = x + 3 or x = -3
at point (-3, 0)

6. Example 26 - (other way round)
Find the point on the curve y = x2 - 6x + 5 where the gradient of the tangent is 14.
gradient of tangent = gradient of curve
= dy/dx
= 2x - 6
so 2x - 6 = 14
or 2x = 20
or x = 10
Put x = 10 into y = x2 - 6x + 5
Point is (10,45)
Giving y =
= 45