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Calculus Made Easy

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

${\begin{cases}y&=x^{2}-5x+6,..................(i)\\y&=ax+b,.......................(ii)\\-1&=2a+b,.......................(iii)\\a&=2x-5,.......................(iv)\end{cases}}$

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

$x^{2}-5x+6=(2x-5)x-1-2(2x-5)$ ,

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.

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