2019 May TZ1 Paper 2 - Q1

We have:

\displaystyle \frac {\mathrm dy} {\mathrm dx} = e^{2x} + x(2e^{2x})

by the product rule.

So, at the point (1, e^2) we have:

\displaystyle \frac {\mathrm dy} {\mathrm dx} = e^2 + 2e^2 = 3e^2

So the equation of the tangent is:

y - e^2 = 3e^2 (x - 1)

When the tangent meets the x-axis, we have y = 0 so:

-e^2 = 3e^2(x - 1)

So:

-1 = 3(x - 1)

so:

\displaystyle x = 1 - \frac 1 3 = \frac 2 3

That is, the coordinates of the intersection point is \displaystyle \left(\frac 2 3, 0\right).