a) 120 b) 20 c) 1 d) 20 e) 1 f) 190 g) 190
$$\begin{align} \text{Suppose } f(x)=-(x-2)^{2}(x+1) \\ \text{Then: } f(x)=-(x^{2}-4x+4)(x+1) \\ =-(x^{3}-4x^{2}+4x+x^{2}-4x+4) \\ = -(x^{3}-3x^{2}+0x+4) \\ = -x^{3}+3x^{2}+4 \\ \therefore f'(x) = -3x^{2}+6x \\ \therefore \text{TP is where f'(x)=0} \\ \therefore x=0,2 \\ \therefore \text{Is TP is where the x intercept double ups?} \end{align} $$ $$\begin{align} \text{Suppose } f(x)=a(x-b)^{2}(x-c) \\ \therefore f(x)=a(x^{2}-2bx+b^{2})(x-c) \\ =a(x^{3}-2bx^{2}+b^{2}x-cx^{2}+2bcx-b^{2}c) \\ = ax^{3}-2abx^{2}+ab^{2}x-acx^{2}+2abcx-ab^{2}c \\ \therefore f'(x)=3ax^{2}-4abx+ab^{2}-2acx+2abc \\ = 3ax^{2}-(2ac-4ab)x+(ab^{2}+2abc) \\ \therefore x=\frac{2ac-4ab \pm \sqrt{ (2ac-4ab)^{2}-4\times3a\times (ab^{2}+2abc) } }{6a} \\ =\frac{2ac-4ab \pm \sqrt{ 4 a^2 b^2 - 40 a^2 b c + 4 a^2 c^2 } }{6a} \\ \text{ blah blah input into symbolab stuff} \\ \text{It follows }x=b,\frac{b+2c}{3} \\ \text{Hence, the gradient of f(x) is 0 when x = b} \\ \text{Thus, x=b is the TP!} \end{align} $$ $$\begin{align} f(x) = x^{3}-4x^{2}+x+6 \\ \therefore f'(x) = 3x^{2}-8x+1 \\ x=\frac{--8\pm\sqrt{ (-8)^{2}-4\times 3 \times 1 }}{2} \\ x=4 + \frac{\pm\sqrt{ 52 }}{2} \\ x= 4+ \frac{\pm 2\sqrt{ 13 }}{2} \\ \therefore x = 4 \pm \sqrt{ 13 } \\ \text{It follows that the turning point is at }x=4-\sqrt{ 13 }, \text{ NOT }x=0\text{ Taj!!} \end{align} $$
