1. Analysing a function near a point on its graph.
Observation: Use the window of your graphing calculator to zoom in
on a point on the graph of a function. What do you see?

2. Question: Is there a line that the graph ‘looks most like’ near
a given point on the graph?

3. Not a definition:
The tangent line to a graph at a point on the graph is the line that
‘looks most like the graph up close’.
We can work this out geometrically for a circle:

4. Problem: Find the equation for the tangent line to the graph of y = x^2
at the point (2,4).
Method: 1 Introduce a change of coordinates x = 2+u, y = 4+v
2 Get the new equation for y = x^2 in terms of u and v.
3 Note that when u and v are small you are close to the point (x,y)=(2,4)
and u^2 and v^2 and u*v are all much smaller than u and v, so up
close to (2,4) we can ignore these terms. What’s left is linear in u and v.
4 Change back to the (x,y) coordinate system using u = x-2, v = y-4
5 Get the equation for the tangent line to y = x^2 at (x,y)=(2,4).

5. Problem:
Use the ‘change of coordinates’ method to recalculate the tangent
line to x^2 + y^2 = 25 at (4,3).

6. Common #2 for Thursday: Find the equation of the tangent line
To the graph of x^2+4x y^2 + 30y = 0 at (-4,1)

7. Common #6
Problem: Find the tangent line to the graph of y = -3x + 4 at the
point (-5,11)
Solution: Before you solve this by the ‘change of coordinates’ method, what do you think the answer should be?

8. The derivative of a function y = f(x) at a point (a,f(a)) on the graph
of the function is the slope f ’(a) of the tangent line to the graph
at (a,f(a)). So we can write the equation for the tangent line at
(a,f(a)) in point slope form: y = f(a) + f ‘(a)(x-a)
Problem: Find the derivative of f(x) = x^2 + 3x at x = a.
Problem: Find the tangent line to the graph of f(x) = x^2 + 3x at
(1,4).