For the graph of y=4+3x−x², determine a) The coordinates of the point where the curve cuts the y-axis b) The coordinates of the point where the curve cuts the x-axis c) The coordinates of any minimum or maximum points d) Hence sketch the graph of y=4+3x−x².

$$x^2+x-2 = (x+2)(x-1)$$

So, the expression becomes:

$$A(x) = \frac{4x+9}{(x+2)(x-1)}$$

To find $P$, let $x = -2$:

$$4(-2)+9 = P(-2-1) + Q(-2+2)$$ $$-8+9 = -3P + 0$$ $$1 = -3P \implies P = -\frac{1}{3}$$

To find $Q$, let $x = 1$:

$$4(1)+9 = P(1-1) + Q(1+2)$$ $$4+9 = 0 + 3Q$$ $$13 = 3Q \implies Q = \frac{13}{3}$$

$$A(x) = \frac{-\frac{1}{3}}{x+2} + \frac{\frac{13}{3}}{x-1}$$ $$\boxed{A(x) = -\frac{1}{3(x+2)} + \frac{13}{3(x-1)}}$$

Question 3b) In a chemical concentration experiment, concentration is modelled by $C(x) = \frac{2x^2+7x+3}{(x+1)^2(x+2)}$. Express $C(x)$ as a sum of partial fractions.

To find $S$, let $x = -1$:

$$2(-1)^2+7(-1)+3 = R(-1+1)(-1+2) + S(-1+2) + T(-1+1)^2$$ $$2 - 7 + 3 = 0 + S(1) + 0$$ $$-2 = S$$

To find $T$, let $x = -2$:

$$2(-2)^2+7(-2)+3 = R(-2+1)(-2+2) + S(-2+2) + T(-2+1)^2$$ $$2(4) - 14 + 3 = 0 + 0 + T(-1)^2$$ $$8 - 14 + 3 = T(1)$$ $$-3 = T$$

To find $R$, let $x = 0$:

$$2(0)^2+7(0)+3 = R(0+1)(0+2) + S(0+2) + T(0+1)^2$$ $$3 = R(1)(2) + S(2) + T(1)^2$$ $$3 = 2R + 2S + T$$

$$3 = 2R + 2(-2) + (-3)$$ $$3 = 2R - 4 - 3$$ $$3 = 2R - 7$$ $$10 = 2R$$ $$R = 5$$

$$C(x) = \frac{5}{x+1} + \frac{-2}{(x+1)^2} + \frac{-3}{x+2}$$ $$\boxed{C(x) = \frac{5}{x+1} - \frac{2}{(x+1)^2} - \frac{3}{x+2}}$$