I'm wondering if someone could help me solve a little conundrum I've been having. I'm trying to find

anginih86

anginih86

Answered question

2022-06-10

I'm wondering if someone could help me solve a little conundrum I've been having. I'm trying to find the equation of a line with slope −1 tangent to the curve of a rational function. The function is y = 1 / ( x 1 ) and the general equation of the line would be y = x + k. I understand that if I set the y-values equal -- that is, 1 / ( x 1 ) = x + k should be able to reorder it into a quadratic and solve for k, giving me the equation of the line. So far, the quadratic I'm finding is y = x 2 + x + k x k 1, yet I can't seem to find a satisfactory value for k that would give me a tangent to the curve. Any thoughts?

Answer & Explanation

stigliy0

stigliy0

Beginner2022-06-11Added 21 answers

The approach you're following in your question is correct, except that last equation should have 0 instead of y i.e. the quadratic equation x 2 + ( k + 1 ) x ( k + 1 ) = 0. The quadratic formula immediately yields roots
x = 1 2 ( ( k + 1 ) ± ( k + 1 ) 2 4 ( k + 1 ) ) = 1 2 ( k + 1 ± k 2 2 k 3 )
Note that the type of roots is determined by the sign of discriminant (i.e. the argument of the square root) and we have two coinciding roots when it vanishes. So we solve
k 2 2 k 3 = 0 for k=−1,3 by inspection. From this we deduce three cases:
1. 1 < k < 3: The discriminant is negative and so there are no real roots
2. k > 3 or k < 1: The discriminant is positive and so there are two real roots
3. k = 1 , 3: The discriminant is zero, so there is exactly one real root
All that remains is to interpret these three cases geometrically...

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