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| *If the curve can be differentiated at any given point, you can obtain the gradient at the given point and thence the formula of the straight line that is perpendicular to it that point. Then you calculate where that line intersects the x-axis. And the rest should be simple. No?*
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In other words, suppose the differentiable curve is y=f(x) and the point of intersection is (x0, f(x0))

The gradient the point of intersection is f'(x0)

The line perpendicular at that point is

y = f(x0) - (x-x0)/f'(x0)

resolving x for y=0 gives

x = f(x0)f'(x0)+x0

if I haven't slipped up somewhere

Bonus assignment: calculate it using vectors rather than coordinates. It's actually even simpler that way.