You are on the right track...you have:
Multiply through by , write the resulting quadratic in standard form, and then use the condition on the discriminant for one repeated root (do you see why?). You will find two values for .
This is for my advanced functions course so NO CALCULUS is used. Please help find the equation.
Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1).
This is my work until I get stuck.
y=-1x+k
-1x+k=1/(x-1)
y=??
I get stuck from here. Can someone please solve this completely and be exact with the steps. Thanks!
You are on the right track...you have:
Multiply through by , write the resulting quadratic in standard form, and then use the condition on the discriminant for one repeated root (do you see why?). You will find two values for .
Well that's what I was getting confused about. I ended up cross multiplying like you're suggesting. But then I don't know if the quadratic is correct, because it certainly looks wrong. And then the discriminant looks even worse.
(k-x)(x-1)-1=0
x^2+x(k+1)+(k-1)=0
Is that the correct quadratic?
I'm not certain how to get a value for k from the discriminant of this? I'm assuming the k doesn't equal 1.
For the quadratic the discriminant is defined to be . When the discriminant is zero, the there is only 1 repeated root for the quadratic. So, equating the discriminant in this case to zero, we find:
Thus, we find two possible tangent lines: