find an equation of the tangent line to the curve at the given point. graph the curve and the tangent line y=x/x-1 at (2,2) )
y = x/(x-1) = 1 + 1/(x-1)
y' = -1/(x-1)^2
y'(2) = -1
so, the tangent line is
y-2 = -1(x-2)
see the graphs at
http://www.wolframalpha.com/input/?i=ploy+y%3Dx%2F%28x-1%29%2C+y%3D-%28x-2%29%2B2
thanks. what rule did you use to find it?
find an equation of the tangent line to the curve at the given point. graph the curve and the tangent line x-3^1/2 at (1,2)
it is suppose to be square root of x-3
find y' at the given point (h,k).
That is the slope of the tangent line.
And then just plug in that slope (call it m) into the point-slope form of the line:
y-k = m(x-h)
Don't forget your Algebra I now that your'e doing calculus. So, what do you get for
y=√(x-3) at (1,2)?
To find the equation of the tangent line to a curve at a given point, you will need to find the derivative of the curve and then substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. After that, you can use the point-slope form of a linear equation to find the equation of the tangent line.
Let's start with graphing the curve and the tangent line.
First, let's graph the curve 𝑦 = 𝑥/(𝑥-1):
To do this, we can choose various x-values, calculate the corresponding y-values, and plot the points on a graph.
For example, if we choose x = -2, 0, 1, 2, and 3, we can calculate the corresponding y-values as follows:
For x = -2:
y = (-2)/((-2)-1) = (-2)/(-3) = 2/3
For x = 0:
y = 0/(0-1) = 0/(-1) = 0
For x = 1:
y = 1/(1-1) = 1/0 (undefined)
For x = 2:
y = 2/(2-1) = 2/1 = 2
For x = 3:
y = 3/(3-1) = 3/2
Plotting these points on a graph, we can connect them to visualize the curve.
Now, let's find the equation of the tangent line at the point (2,2).
To find the derivative of the curve 𝑦 = 𝑥/(𝑥-1), we can use the quotient rule. The derivative is given by:
𝑦' = [ (𝑥-1) * 𝑑(𝑥) - 𝑥 * 𝑑(𝑥-1) ] / (𝑥-1)^2
Simplifying further, we have:
𝑦' = [ (𝑥-1) * 1 - 𝑥 * (-1) ] / (𝑥-1)^2
= [ 𝑥 - 1 + 𝑥 ] / (𝑥-1)^2
= (2𝑥 - 1) / (𝑥-1)^2
Now, we substitute the x-coordinate of the given point (2,2) into the derivative to find the slope of the tangent line:
𝑦' = (2(2) - 1) / (2-1)^2
= (4 - 1) / 1
= 3
The slope of the tangent line at (2,2) is 3.
Using the point-slope form, we can write the equation of the tangent line:
𝑦 - 2 = 3(𝑥 - 2)
Simplifying further, we have:
𝑦 - 2 = 3𝑥 - 6
𝑦 = 3𝑥 - 4
Therefore, the equation of the tangent line to the curve 𝑦 = 𝑥/(𝑥-1) at (2,2) is 𝑦 = 3𝑥 - 4.
Now, we can plot the tangent line on the same graph as the curve, and we have successfully found the equation of the tangent line.