Page:Calculus Made Easy.pdf/110

 The slope of the tangent must be the same as the $$\dfrac{dy}{dx}$$ of the curve; that is, $$2x-5$$.

The equation of the straight line is $$y=ax+b$$, and as it is satisfied for the values $$x=2$$, $$y=-1$$, then $$-1 = a\times 2 + b$$; also, its $$\dfrac{dy}{dx}=a=2x-5$$.

The $$x$$ and the $$y$$ of the point of contact must also satisfy both the equation of the tangent and the equation of the curve.

We have then

four equations in $$a$$, $$b$$, $$x$$, $$y$$.

Equations (i) and (ii) give $$x^2-5x+6=ax+b$$.

Replacing $$a$$ and $$b$$ by their value in this, we get

which simplifies to $$x^2-4x+3=0$$, the solutions of which are: $$x=3$$ and $$x=1$$. Replacing in (i), we get $$y=0$$ and $$y=2$$ respectively; the two points of contact are then $$x=1$$, $$y=2$$; and $$x=3$$, $$y=0$$.

Note.—In all exercises dealing with curves, students will find it extremely instructive to verify the deductions obtained by actually plotting the curves.