Well, we got our tests back today and I did better than I thought I did, but missed questions I thought I’d gotten right and got questions right I thought I had done wrong. That’s kind of frustrating. So today will be dedicated to my biggest mistake, the one I corrected last time and yet STILL have not learned from. I STILL cannot find the stupid equation of the line that is tangent to the function. So I am going to put the problem up here, take myself through it step-by-step, and hopefully make it stick this time.

Find the equation of the line that is tangent to the function f(x)=x^2+6x-1 at the point (1,6).

1. First, you are going to want to use point-slope form, something I continually forget the equation for. y-y_1=m(x-x_1)

2. Then you want to solve for m by finding the derivative of F and plugging in the x-coordinate.

f(x)=x^2+6x-1 f'(x)=2x+6 if x=1, 2(1)+6 = 8 so m=8

3. y-6=8(x-1) = y=8x-8+6 = y=8x-2

It’s not hard. I just goofed. I have no idea how I got this wrong, but I managed to do so, and I don’t plan on letting myself mess it up again.

 

I hope you all had a good test score, and to sweeten my post, I wanted to add a little chart I made from the notes we did in class today:

Functions Tells us what about F When Positive When Negative When zero
F (original Function) y-coordinate Graph is above x-axis in either quadrant I or II Graph is below x-axis in either quadrant III or IV On the x-axis (x-intercepts/roots)
F’ (first derivative) -slope of the tangent line

-instantaneous rate of change

-limit of the difference quotient

-whether graph of F is increasing or decreasing

 Graph of F is increasing Graph of F is decreasing -critical points

-might be a relative extremum

-horizontal tangent line
F’’ (2nd derivative) Concavity of F -The graph of F is concave up

– F’ is increasing

 – The graph of F is concave down

– F’ is decreasing

 – Might be inflection point (check that it actually changes concavity)