GROVE PRELIM 2U/3U TEST YOURSELF 8 - Q20

$y'={vu'-uv' \over v^2}$

$y'={(2x+1)4-(4x-3)2 \over (2x+1)^2}$

Put $x=1$

$y'={(2(1)+1)4-(4(1)-3)2 \over (2(1)+1)^2}$

$y'=\displaystyle{10\over 9}$

When $x=1$, $y=(4-3)/(2+1)=1/3$.

$(1, 1/3)$.

$y-{1\over 3}={10\over 9}(x-1)$

$9y-3=10(x-1)$

$9y-3=10x-10$

$10x-9y-7=0$

When $y=0$,

$10x-7=0$

$x=\displaystyle {7\over 10}$