ap calculus question..help?

Question

well, im currently learning about derivatives and everything dealing with derivatives... what is the process to find the eqn. of the tangent line given f(x).. please provide example

Answer 1

It would be easy to draw a parabola.
One point along this parabola is: (x,f(x))
Another point along this parabola is: (x + h,f(x + h))
The slope between these two points is m...
m = (y2 - y1)/(x2 - x1)
m = ( f(x + h) - f(x) )/( x + h - x )
m = ( f(x + h) - f(x) )/( h )
If you would notice as h approaches zero, the two points become closer to one another and a tangent seems to be produced.
When h = 0, you have a tangent slope!
f'(x) = lim[ h~> 0 ] ( f(x + h) - f(x) )/( h )

Answer 2

Given f(x), determine f'(x) - the derivative of your original given equation. Once you have f'(x):

*select an x-value at which you want to determine the equation of the tangent line
*determine f(x) at your selected x-value, so that you have the coordinates of a point on the original curve (x, f(x))
*determine f'(x) at your selected x-value; this is the slope of the curve f(x) at your selected point (x, f(x)) on the original curve
*now you have both the slope of the tangent line and a point (x, f(x)) on that line, so determining the equation of the tangent line should be a breeze for an algebra survivor like yourself

Example: f(x) = x^2 - 6x - 16; what is the equation of the tangent line at x = 2?

f(x) = x^2 - 6x - 16
f(3) = 2^2 - 6(2) - 16 = 4 - 12 - 16 = -24
f'(x) = 2x - 6
f'(3) = 2(2) - 6 = 4 - 6 = -2

The y-intercept of the line that contains the point (2, -24) and has a slope of -2 is calculated as follows:

-24 = (-2)*2 + b
-24 = 4 + b
-28 = b

The equation of the tangent line is y = -2x - 28

Answer 3

Let (a,b) be the point of tangency.
Then the eqn of the line(in point slope form)
is y - b = m(x-a).
Now, how do you find m, the slope?
You find the derivative at x = a. So
your tangent line has equation y-b = f'(a)(x-a).
Example: What is the equation of the tangent line to y = x³
at the point(1,1)?
It is y - 1 = f'(1)(x-1).
But f'(x) = 3x², so f'(1) = 3.
Answer y-1 = 3(x-1)
or y = 3x-2.
You will meet this idea many times in your course.

Answer 4

Nothing against you, but yor calculus class is waaaaay behind if this is what you are currently learning about.


This is what you are asking for:

1) You start with a specific x value (let's call it c) and a function f(x).

2) You want the point on the graph where x = c, so find f(c). Your ordered pair is (c, f(c)).

3) You want the slope of the tangent line. Find the derivative f'(x), then plug in c. Your slope is f'(c).

4) Use the point (c, f(c)) and the slope f'(c) to write an equation in point-slope form. Point-slope form is

y - y1 = m(x - x1),

where m is the slope and (x1, y1) is the point. In this case,
m = f'(c), x1 = c, and y1 = f(c).


5) If desired, convert to slope-intercept form by solving or y.


As mentioed above, this is pretty basic stuff. Especially this far into a calculus class.

Answer 5

if function f(x)
the process is the following steps
1. u have 1 of 2 things either a point (x1,y1) and u want to find the tangent at this point or u may have only x1 so u need to plug the value x1in f(x) to get y1

2. get f' (x)
3. plug x1 in f'(x) and get the value of f'(x1) ----> this is the slope of tangent (m)
4. u now have an equation of the tangent as
y= m x +b = f'(x1) x +b
5. to get b plug the point (x1,y1) into the equation
y1= f'(x1) x1+ b
b= y1 - f'(x1) x1
6. so the equation will be
y= f'(x1) x + y1 -f'(x1) x1

Answer 6

Well I'm sure the text doesn't have any examples of one of the core concepts of calculus....they just gave it to you so you could fund the chiropractor's for lugging it around. READ THE BOOK FOR EXAMPLES.

You could also find good examples online, Paul gives great notes here with examples and all!

http://tutorial.math.lamar.edu/classes/calcI/tangents_rates.aspx