Question

Equation of the tangent to the curve $y(x-2)(x-3)-x+7=0$ at the point where
the curve cuts x-axis is

• A. $x-20y=7$
• B. $x-20y+7=0$
• C. $x+20y-7=0$
• D. $x+20y+7=0$

Answer 1

The correct option is A $x-20y=7$

$y=0$ (at x-axis)
$0-x+7=0$
$x=7$
$y=\frac{x-7}{(x-2)(x-3)}$
$y^{\prime }=\frac{(x-2)(x-3)-(x-7)(2x-5)}{(x-2{)}^2(x-3{)}^2}$
put $x=7$
$\frac{5\times 4-\left(0\right)}{(5\times 4{)}^2}$
$y^{\prime }=\frac{1}{20}$
$(y-0)=\frac{1}{20}(x-7)$
$x-20y-7=0$