1. Starter
The line L passes through the points (0, 7) and
(3, 19). Work out the equation of the line L.
2. Starter
The line L passes through the points (0, 7) and
(3, 19). Work out the equation of the line L.
Gradient =
19−7
3−0
=
12
3
= 4
Equation: y=4x+c
Passes through (0,7) so y-intercept is 7
Therefore the equation is y=4x+7
How would we have worked out c if we were not given the y-intercept?
3. Calculus - Differentiation
• Differentiation is a way of finding a gradient at
a point on a curve.
• It is needed as curves have (by definition) a
constantly changing gradient.
4. Differentiation
• Why and how differentiation works is not
required knowledge for the Further Maths
exam
• It will not be covered in this session (look up
Differentiation from First Principles if you
want some light summer reading)
5. Differentiation – How To…
OR:
Multiply the whole thing by the power and
reduce it by one
𝑦 = 𝑎𝑥 𝑛
𝑑𝑦
𝑑𝑥
= a𝑥 𝑛−1
9. Differentiation - Tangents
A tangent is a line that
touches a curve at a single
point.
The gradient of the
tangent is equal to the
gradient of the curve at
that point.
10. Differentiation - Tangents
As a straight line the
equation of the tangent is:
y=mx + c
This is equal to evaluated at the point P.
𝒅𝒚
𝒅𝒙
You will need to know a specific point on the
line to find c.
15. Tangents and Normals
• A normal is a line that is perpendicular to the
tangent at a specific point.
• The gradient of a normal is the negative
reciprocal of the tangent (-1/tangent gradient)
16. Tangents and Normals
Tangent Gradient Normal Gradient
4 -1/4
-3 1/3
1/2 -2
-1/3 3
3/4 -4/3
-7/2 2/7
Examples of negative
reciprocals
Once you have worked
out the gradient, finding
the equation is exactly the
same as for the tangent