The line $y=5x-3$ is a tangent to the curve $y=kx^2-3x+5$ at the point $A$. Find the value of $k$.

Question

So, the derivative of the function should be equal to the gradient of the tangent line: $5= 2kx-3$.

But I don't know what I should do next. Any hints please?

Answer 1

Not only does $y=5x-3$ have to have the same slope as $y=kx^2-3x+5$ at the point $A$, but both curves must contain the point $A$. If $A=(a,b)$, this tells you that $5=2ka-3$ (this is the condition you found at the point $(x,y)=(a,b)$) and also $5a-3=ka^2-3a+5$. Now use both equations to solve for $k$.

Answer 2

1. The slope of the tangent line and the actual tangent should be the same in point A.
2. The slope of the curve is $2kx -3$
3. Equality in $A$

Answer 3

I presume you aren't given point A so you can't just plug that in for $x$ and figure out k, so you could set up a system of two equations with two unknowns, $x$ and $k$. The first equation can be gotten from how you derived $5 = 2kx-3$ and the second can be gotten by equating both equations such as to get $$ 5x-3=kx^2-3x+5$$ You could then go about solving for $x$ and $k$

Answer 4

Set $$5x-3=kx^2-3x+5\implies kx^2-8x+8=0$$

This quadratic must have double roots since the line is tangent to the curve.

Therefore $$64-4\times k\times 8=0$$

So $$k=2$$