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?

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

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:

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).

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

Common #2 for Thursday: Find the equation of the tangent line To the graph of x^2+4x -27 -3y^2 + 30y = 0 at (-4,1)

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?

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).