(a) $\frac{x^2-4}{x^2+x}÷\frac{x^2-4x+4}{x+1}$
= $\frac{x^2-4}{x^2+x}\times \frac{x+1}{x^2-4x+4}$
= $\frac{(x-2)(x+2)}{x(x+1)}\times \frac{x+1}{x^2-2x-2x+4}$ (Using difference of two squares)
= $\frac{(x-2)(x+2)}{x}\times \frac{1}{(x-2{)}^2}$
= $\frac{(x+2)}{x(x-2)}.(b)Rootsof\text{(}a{x}^2+bx+c=mx+k$
For graph $y=a{x}^2+bx+c$,
$x=-1$ and $x=2$
$x+1=0$ and $x-2=0$
$(x+1)(x-2)=0$
$x^2-2x+x-2=0$
$x^2-x-2=0.....(i)$
$-x^2+x+2=0......(ii)$
For graph $y=mx+k$
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-1)}{1.3-(-1.3)}$
$m=\frac{2}{2.6}=0.769$
$k=0.25$
$x^2-x-2=0.769x+0.25$
$x^2-x-0.769x-2-0.25=0$
$x^2-1.769x-2.25=0$
Using formula method, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-(-1.769)\pm \sqrt{(-1.769{)}^2-4(1)(-2.25)}}{2(1)}$
$x=\frac{1.769\pm \sqrt{3.1294+9}}{2}$
$x=\frac{1.769\pm 3.483}{2}$
$x=\frac{1.769+3.483}{2}$ or $x=\frac{1.769-3.483}{2}$
$x=\frac{5.252}{2}\approxeq 2.626$ or $x=\frac{-1.714}{2}\approxeq -0.857$
(ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively.
Using coordinate for L:
$0=a(-1^2)+b(-1)+c⟹0=a-b+c.....(1)$
Using M :
$2=a(0^2)+b(0)+c⟹c=2......(2)$
Using N :
$0=a(2^2)+b(2)+c⟹0=4a+2b+c....(3)$
From (2), c = 2
$\therefore (1):a-b+2=0⟹a-b=-2....(1a)$
$\therefore (3):4a+2b+2=0⟹4a+2b=-2.....(3a)$
Solving equations (1a) and (3a) simultaneously, we get
$a=-1,b=1,c=2$
$\therefore$ Equation of the curve : $-x^2+x+2=0$
(iii) Line of symmetry = $x=\frac{-b}{2a}$
$y=-x^2+x+2$
Line of symmetry = $\frac{-1}{2(-1)}$
$x=\frac{1}{2}$