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

calculustangent line

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?

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 $k$.