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

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SUMMARY

The discussion centers on finding the equation of the tangent line to the curve defined by the function y=2x-x^2 at the point (2, -4). The correct tangent line equation is derived as y + 4 = -10(x - 2), where the slope is determined by differentiating the function to obtain dy/dx = 2 - 2x. The participants clarify the importance of understanding derivatives, as they represent the slope of the tangent line at a given point. Additionally, a correction was noted regarding the function, which was initially misrepresented as y=2x-x^3.

PREREQUISITES
  • Understanding of basic calculus concepts, specifically derivatives.
  • Familiarity with the equation of a line in slope-intercept form (y = mx + b).
  • Ability to differentiate polynomial functions.
  • Knowledge of evaluating functions at specific points.
NEXT STEPS
  • Learn the rules of differentiation for polynomial functions.
  • Study the concept of limits as they relate to derivatives.
  • Practice finding tangent lines for various functions using the derivative.
  • Explore the application of derivatives in real-world scenarios, such as optimization problems.
USEFUL FOR

Students preparing for precalculus or calculus exams, educators teaching derivatives, and anyone seeking to understand the application of derivatives in finding tangent lines to curves.

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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The slop of the tangent can be found by differentiating the given equation, let it be \frac {\delta y}{\delta x}.

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

where, x_1 and y_1 are the given points!

Substitute, and you'll get the answer, rightaway! :wink:
 
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.
 
Thanks, but could somebody explain basic derivatives in general, please?
 
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.
 
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).
 
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,

y=2x - x^3
\frac{\delta y}{\delta x}=~ 2-3x^2

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
(y-y_1)=\frac{\delta y}{\delta x} (x-x_1)

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.
 

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