[Highschool - Calculus] Finding Equation of a Tangent Line and Normal Line

[180digi]

Currently, we are on chapter three: Derivatives. Here is the problem.

1) If f(2)=3 and (f)'(2)=5, find an equation of a) the tangent line, and b) the normal line to the graph of y=f(x) at the point where x=2.

I have no idea how to approach this problem, and neither my notes nor the section examples give any help. I may be in over my head taking AP Calculus AB, however I took it because highschool statistics is a joke and there were no college prep highschool calculus courses to be offered at my school. Thank you.

[Jtehanonymous]

f(2) = 3 gives you the points on the line.

f'(2) = 5 gives you the slope at point two of the tangent line. From this, you can write the slope of the tangent line.

x = 2
y = 3
m = 5

Point-slope form:

y - y(1) = m(x - x(1))

y - 3 = 5(x - 2)


You can put that in slope-intercept form if you like.

Next, for the normal line, the slope of a normal line is just the opposite reciprocal of the tangent line's slope. So therefore your slope just changes to -1/5

So....again...

y - 3 = -1/5(x - 2)

And again, you can put that in slope-intercept form if you like.

Just remember that derivative is another word for "slope". So the f'(2) = 5 can really be read, "the slope at 2 is 5"

Hope this helps.

[180digi]

Thank you. I'd like to pose another question instead of another topic.

3) Use the definition f'(a) = the limit as x->a f(x)-f(a)/x-a to find the derivative of f(x) = 1/x at x=3.

I realize you start with the definition, but where do you get x and a from? 3?

[Relambrien]

The x is just the variable, and a comes from the information you're given.

f(x) = 1/x at x=3. So the "a" you're looking for is 3. It's kind of confusing considering it says x=3 rather than a=3, but what they're trying to get across is this: if you ever see just the variable x, it refers to any point along the x-axis. A value "a" or whatever will be a specific point along that axis.

So you're looking to solve this:

lim (1/x - 1/3)/x-3
x>3

Your goal is going to be to rewrite the numerator in such a way as to be able to cancel something out with the denominator, so that when you plug 3 in for x, the denominator doesn't go to zero.

[zhul4nder]

well first derive the equation, which would be 1/x^2 correct? Then you just pug in the three and get your answer.
1/x^2 = 1/3^2 = 1/9 ?

[dag12]

3) Use the definition f'(a) = the limit as x->a f(x)-f(a)/x-a to find the derivative of f(x) = 1/x at x=3.

so.
lim [x->a] [ f(x) - f(a) ] / (x - a)
lim [x->a] [ 1/x - 1/a ] / (x - a)
lim [x->a] [ (a - x) / ax ] / (x - a)
lim [x->a] [ - (x - a) / ax ] / (x - a)
lim [x->a] - 1 / ax
= - 1 / x^2

f'(x) = - 1 / x^2

So when x = 3,
f'(3) = - 1/9