I don't understand how to do problems like the following:
The equation of the line tangent to the curve y=(kx+8)/(k+x) at x=-2 is y=x+4. What is the value of k?
Thanks :-)
The equation of the line tangent to the curve y=(kx+8)/(k+x) at x=-2 is y=x+4. What is the value of k?
Thanks :-)
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when x= -2, the y value of the tangent is
y= -2+4 =2
Since it is a tangent to the curve, the curve must have the exact same y value at x=-2 (because that is the definition of a tangent, it intersests at one common point)
Therefore, (-2,2) lies on the curve
Subbing into the equation of the curve
2= (-2k+8)/ (k-2)
2(k-2)= -2k+8
2k-4 = -2k +8
4k = 12
k=3
No derivatives needed to be calculated at all.
y= -2+4 =2
Since it is a tangent to the curve, the curve must have the exact same y value at x=-2 (because that is the definition of a tangent, it intersests at one common point)
Therefore, (-2,2) lies on the curve
Subbing into the equation of the curve
2= (-2k+8)/ (k-2)
2(k-2)= -2k+8
2k-4 = -2k +8
4k = 12
k=3
No derivatives needed to be calculated at all.
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"Math" showed that k=3 just using the point (-2, 2).
You can also find this using the derivative: y' = [ (k+x)(k) - (kx+8)(1) ] / (k+x)^2 = (k^2 - 8) / (k+x)^2. Since the tangent line at x=-2 has slope 1, we expect y'(-2) = 1.
That tells us k^2 - 8 = (k-2)^2, which means k^2 - 8 = k^2 - 4k + 4, so 4k = 12, and k=3.
I like "Math"'s way better!
You can also find this using the derivative: y' = [ (k+x)(k) - (kx+8)(1) ] / (k+x)^2 = (k^2 - 8) / (k+x)^2. Since the tangent line at x=-2 has slope 1, we expect y'(-2) = 1.
That tells us k^2 - 8 = (k-2)^2, which means k^2 - 8 = k^2 - 4k + 4, so 4k = 12, and k=3.
I like "Math"'s way better!
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Due to the voluminous amount of fan mail Krusty receives , he is unable to answer your question. keep watchin' kids!