Find Equation of Tangent Line to y=2x-x^2 at (2,-4)

In summary: Your earlier post made it seem like (y - y1)/(x - x1) was the derivative, which it isn't.In summary, to find the equation of a tangent line to a curve at a given point, you need to differentiate the equation of the curve to find the slope of the tangent line, then use the point given to find the equation of the tangent line using the formula (y-y1)=m(x-x1), where m is the slope of the tangent line.
  • #1
jacksonpeeble
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Homework Statement


Find an equation of the tangent line to the curve y=2x-x^2 at the point (2, -4).


Homework Equations


y=2x-x^2


The Attempt at a Solution


I do not have any idea how to do any derivative, other than the equation to do so. Please note that we have been given the answer key - I already know that the solution is y+4=-10(x-2).

This is for a Precalculus exam (but I've been informed that derivatives are calculus). I have been absent for a while, which is when we learned about derivatives, so any background information would be nice as well, but the test is tomorrow.
 
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  • #2
The slop of the tangent can be found by differentiating the given equation, let it be [tex]\frac {\delta y}{\delta x}.[/tex]

Now, the formula to find the equation is simply,
[tex]\frac{y-y_1}{x-x_1}=~\frac {\delta y}{\delta x}.[/tex]

where, [tex] x_1[/tex] and [tex] y_1[/tex] are the given points!

Substitute, and you'll get the answer, rightaway! :wink:
 
  • #3
That's only for equations of the form y = m*x + b i.e. linear for some form of linear. In general dy/dx is not that.
 
  • #4
Thanks, but could somebody explain basic derivatives in general, please?
 
  • #6
Now I am ;-). I'll post back if I'm still confused. It would be really great if (if anyone has the time) somebody could post a walk-through on an example problem, though.
 
  • #8
If anyone is still following this thread: after checking with other students in the class, I found that the teacher had corrected the problem on the sheet in class - f(x) was supposed to equal 2x-x^3, which makes a major difference.

NoMoreExams - I understand that you're trying to be helpful, and I assume that you must get people that don't bother searching elsewhere before posting, but I do. I searched Google, and had the page up for Tangent Lines on Wikipedia before you provided the Derivatives link. I had even run several advanced searches on Physics Forums to try to find very similar problems to mine.

So even though I may be the exception, I find it frustrating when people tell me to check in other places when I, personally, use posting on a forum only as a last resort (I don't believe in swamping forums with repeats of previous topics).
 
  • #9
But doesn't the 2nd page on the google search I linked give you the necessary tutorial you need to solve your problem?
 
  • #10
The tutorial in NoMoreExams' link does cover the process needed to solve your problem.
 
  • #11
The slop of the tangent can be found by differentiating the given equation, let it be

Now, the formula to find the equation is simply,where, and are the given points!

Substitute, and you'll get the answer, rightaway!

Didnt you try this method? It'll give the answer straightaway,

[tex]y=2x - x^3 [/tex]
[tex]\frac{\delta y}{\delta x}=~ 2-3x^2[/tex]

Now, this is the equation for the slope, substituting value of 'x' you'll get the slope of the line to be -10!
Then have this value in the equation mentioned above to get the equation of the tangent!

P.S:NoMoreExams please note this.:smile:
 
  • #12
What am I noting? That evaluating the derivative at a point will give you the slope of the tangent line? I kind of knew that already. My point was that the derivative is not the same as (y - y1)/(x - x1) as you wrote it.
 
  • #13
[tex] (y-y_1)=\frac{\delta y}{\delta x} (x-x_1) [/tex]

This is the exact representation of the equation of the required equation!
 
  • #14
Once again, that's for a tangent line in which case it's true that the derivative of the tangent line will be its slope. That's what derivative, or one of the things, a derivative represents the slope of your curve at that point i.e. the slope of the tangent line to your curve at a particular point. My point wasn't to say you're flat out wrong, it was to clarify your point.
 

FAQ: Find Equation of Tangent Line to y=2x-x^2 at (2,-4)

What is the equation of the tangent line to y=2x-x^2 at the point (2,-4)?

The equation of the tangent line at a given point on a function is given by y-y1 = m(x-x1), where (x1,y1) is the given point and m is the slope of the tangent line. In this case, the point (2,-4) is on the function y=2x-x^2, and the slope at this point can be found by taking the derivative of the function and plugging in x=2. The derivative of y=2x-x^2 is 2-2x, and when x=2, the slope is 2-2(2) = -2. Therefore, the equation of the tangent line is y-(-4) = -2(x-2), which simplifies to y=-2x+4.

How do you find the slope of the tangent line to y=2x-x^2 at the point (2,-4)?

The slope of the tangent line at a given point on a function can be found by taking the derivative of the function and plugging in the x-coordinate of the given point. In this case, the derivative of y=2x-x^2 is 2-2x, and when x=2, the slope is 2-2(2) = -2.

Can you provide an example of finding the equation of a tangent line to a function at a given point?

Yes, let's say we want to find the equation of the tangent line to the function y=3x^2-2x at the point (1,1). First, we take the derivative of the function, which is 6x-2. Then, we plug in x=1 to find the slope at that point: 6(1)-2 = 4. The equation of the tangent line is y-1 = 4(x-1), which simplifies to y=4x-3.

How does the graph of the tangent line relate to the function at the given point?

The tangent line at a given point on a function is a straight line that touches the function at that point. This means that the value of the function at the given point will be the same as the y-intercept of the tangent line. In other words, the tangent line and the function share a common point at the given point, but the tangent line represents the instantaneous rate of change of the function at that point.

Is it possible for a function to have more than one tangent line at a given point?

Yes, it is possible for a function to have more than one tangent line at a given point. This can happen when the function has a point of inflection, where the slope of the function changes from increasing to decreasing or vice versa. In this case, there will be two tangent lines to the function at that point, each with a different slope.

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