Find Tangent Line: f(2) & f'(2) at y=4x-5

In summary, the conversation discusses finding the value of f(2) and f'(2) for an equation of the tangent line to the curve y=f(x) at the point a=2, where y=4x-5. The initial attempt at solving the problem involved using the slope formula and identifying the point (2,3). The relationship between the slope of the tangent line and the derivative is established, leading to the solution of f'(2) = 4 and f(2) = 3.
  • #1
physics604
92
2
1. If an equation of the tangent line to the curve y=f(x) at the
point a=2 where is y=4x-5, find f(2) and f'(2).


Homework Equations



m=[itex]\frac{f(x)-f(a)}{x-a}[/itex]

The Attempt at a Solution



To be honest, I really don't know where to start. Here's what I have so far:

m=[itex]\frac{f(x)-f(a)}{x-a}[/itex]

I know slope is 4 according to the equation above. Also, I know there is a point (2,3), plugging a into the equation.

4=[itex]\frac{f(x)-3)}{x-2}[/itex]

Now what can I do? This doesn't help me find f(2) or f'(2).

Any help would be greatly appreciated.
 
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  • #2
physics604 said:
To be honest, I really don't know where to start. Here's what I have so far:

m=[itex]\frac{f(x)-f(a)}{x-a}[/itex]

I know slope is 4 according to the equation above.
What is the relationship between the slope of the tangent line and the derivative?
 
  • #3
The slope -is- the derivative.
 
  • #4
OK, good. So you established that the slope of the tangent line at the point ##x=2## is ##4##. What does that tell you about ##f'(2)##?
 
  • #5
I got it! Thanks!

f'(2) = 4 and f(2) = 3!
 
  • #7
physics604 said:
I got it! Thanks!

f'(2) = 4 and f(2) = 3!
Looks good. I'll take a look at your other question now.
 

What is the purpose of finding the tangent line for f(2) & f'(2) at y=4x-5?

The purpose of finding the tangent line is to determine the slope of the curve at a specific point (in this case, x=2) and to find the equation of the line that is tangent to the curve at that point.

How do I find f(2) & f'(2) for the function y=4x-5?

To find f(2), simply substitute x=2 into the given function. So, f(2)=4(2)-5=3. To find f'(2), take the derivative of the function, which is the slope of the tangent line. In this case, f'(x)=4, so f'(2)=4.

What is the equation of the tangent line for f(2) & f'(2) at y=4x-5?

The equation of the tangent line can be found using the point-slope formula, which is y-y1=m(x-x1), where m is the slope (f'(2)) and (x1,y1) is the point (2,3). So, the equation of the tangent line is y-3=4(x-2).

How can I graph the tangent line for f(2) & f'(2) at y=4x-5?

To graph the tangent line, first plot the point (2,3) on the function y=4x-5. Then, use the slope (f'(2)=4) to find another point on the line. In this case, since the slope is positive, you can go up 4 units and right 1 unit from the point (2,3) to get another point on the line. Plot this point and draw a straight line through the two points to represent the tangent line.

Why is finding the tangent line important in calculus?

Finding the tangent line is important in calculus because it allows us to approximate the behavior of a curve at a specific point. This is useful in many real-life applications, such as predicting the speed of an object or determining the rate of change of a function.

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