The equation of a line to a point on a curve (point slope form) includes the slope and the position of that line on that curve function. It’s better than the derivative because the derivative only tells us the slope. Again in Algebra the professor forgot to tell us the importance of that and the relationship to the derivative. Didn’t make it interesting enough to sink in and how it relates to the real world.

You have a function. Has to have x^2 in it to be a curve from my understanding, Example y or f(x) = 3x^2

Graph that and you have some form of curve in this case a “valley” parabola, (my own word), -3x^2 and you have a “mountain” parabola (again my own word).

Pick any point “(x,y)” Example: (3,12)

Point = (3,12)

x = 3

y = 12

Find the derivative: f(x) = 3x^2

f’(x) = (2)(3)x^(2-1)

= 6x^(1)

= 6x

Compute the derivative at the point “x”

f’(3) = 6(3)

= 18 = m = slope

Point slope form = y = mx + b

y = 12 so:

12 = mx + b

m = 18

12 = 18x + b

x = 3 so:

12 = 18(3) + b

= 54 + b

b = 12 – 54

= – 42

y = mx + b

= 18x + -42

If you graph this equation along with the original function you’ll see the tangent line to that point on the curve

The slope = 18/1 (rise over run)

The angle of that tangent line = tan^(-1)(18/1) = 86.8 degrees

(make sure your calculator mode is in APPROXIMATE and DEGREES)

Fabulous and exciting, right? lol Tom

p.s. You’ll love my programs

Have a test or quiz on point slope form? Here is a video example using the programs on the TI-89 calculator: (Click Links below)

Bon says

They don’t cover everything beusace there isn’t enough time in the course schedule to do everything. Rather than get a whirlwind tour of the entire book they prefer to cover in enough detail the sections they think are most relevant to the course topic. Even advanced course levels are still too wide a net to cover all the material within that area in just one semester.