Find the value of k such that the equation of the tangent line to f(x) = x^2 + k*x is equal to y = 5*x + 3
@ganeshie8
\(x^2+kx =5x+3 \implies x^2+x(k-5)-3=0 \) Notice that the discriminant is always \(\ge 12\), therefore \(y=5x+3\) cannot be a tangent
lol I found the tangent line at (a,f(a)) since this was not given...; The tangent line was: \[y=(2x+k)x-a^2 \\ \\ -a^2 \neq 3\]
@ganeshie8 can you explain the part about the discriminant \(\ge 12\)?
mistype for me
\[y=(2\color{red}a+k)x-a^2\]
Ahh nice, \(-a^2\ne 3\) for any real \(a\) so no solutions..
@Jhannybean basically we get tangent line only if below system has a single solution : ``` y = x^2+kx y = 5x+3 ```
that's true for that line to be tangent at some point on the curve that tangent line and curve thing will have to share a common point
In other words, we want both the line and parabola to intersect at just one point since the parabola opens up, the above system must have exactly one solution for tangent line
|dw:1440296346157:dw|
it is easy see from the graph that the tangent lien cannot cut the parabola back, at a different place..
Yeah,the explanation and the graph makes sense, i was just wondering where the 12 came from.
I can't believe this was posted a year ago... I guess we will never be able to ask the asker for any corrections to his problem.
``` y = x^2+kx y = 5x+3 ``` set the right hand sides equal to each other, you get a quadratic the discriminant of the quadratic must have to be 0, but it happens that \(D\ge 12\) so no such tangent
OHHH.... thats what i was missing haha, i get it now!
wow! it was posted one year ago... must be tha the OP has accidentally bumped it..
Maybe they were STILL working on the same problem.....from a year ago!
I don't know... Everything thing he has done was over a year ago. There has not been any other activity in posting answers or questions since a year.
According to the profile.
@medisynergi22
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