(a) Divide \(\frac{x^{2} - 4}{x^{2} + x}\) by \(\frac{x^{2} - 4x + 4}{x + 1}\). (b) The diagram below shows the graphs of \(y = ax^{2} + bx + c\) and \(y = ...

(a) Divide the algebraic fractions.

Dividing by a fraction is the same as multiplying by its reciprocal:

\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]

Factorise each expression fully:

\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]

Substitute the factorised forms:

\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]

Cancel the common factors \((x+1)\) and one \((x-2)\):

\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\]

(b) Using the graph.

Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.

(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)

Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):

\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]

Put \(c=2\) from (2) into (1) and (3):

\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]

From (1a), \(b=a+2\). Substitute into (3a):

\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]

Then \(b=a+2=1\) and \(c=2\). Therefore

\[a=-1,\qquad b=1,\qquad c=2,\]

and the equation of the curve is

\[y=-x^{2}+x+2.\]

(i) Roots of \(ax^{2}+bx+c=mx+k\).

The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):

\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]

Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).

Now equate the curve and the line:

\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]

Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):

\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]

The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.

(iii) Line of symmetry of the curve.

The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):

\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]

The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).