How can i find the equation of a tangent to a curve given an equation and a point

Question

the question says dy/dx = e^kx (k is constant) is such that the tangent at (1,e^2) passes through the origin, find the equation of this tangent. I have subbed in the point to find k=2 (therefore dy/dx=e^2x) however i can not find the equation of the tangent.

help me pls

Answer 1

Hello,

The equation of a tangent line is just a linear line that is tangent to a function at a particular point. So the general equation of a tangent line is the same general equation of a linear line, namely: y = mx + b where m is the slope and b is the y-intercept.

The slope of the tangent line is the same as the slope of the original function at the point, which means all we need to do is plug in the x-value of the tangent point into dy/dx:

dy/dx = e^2x

dy/dx = e²

slope = m = dy/dx = e²

And since the problem states the tangent line passes through the origin, we know the y-intercept is 0. Therefore the equation of our tangent line is:

y = mx + b

y = e²x + 0

y = e²x