Quadratic Functions, SPM Practice (Long Questions)

Question 3:
Given that the quadratic function f(x) = 2x² – px + p has a minimum value of –18 at x = 1.
(a) Find the values of p and q.
(b) With the value of p and q found in (a), find the values of x, where graph f(x) cuts the x-axis.
(c) Hence, sketch the graph of f(x).

Solution:
(a)
$\begin{array}{l}f\left(x\right)=2x^2-px+q\\ =2\left[x^2-\frac{p}{2}x+\frac{q}{2}\right]\\ =2\left[{\left(x+\frac{-p}{4}\right)}^2-{\left(\frac{-p}{4}\right)}^2+\frac{q}{2}\right]\\ =2\left[{\left(x-\frac{p}{4}\right)}^2-\frac{p^2}{16}+\frac{q}{2}\right]\\ =2{\left(x-\frac{p}{4}\right)}^2-\frac{p^2}{8}+q\end{array}$


$\begin{array}{l}\therefore \frac{p}{4}=1------\left(1\right)\\ \text{and }-\frac{p^2}{8}+q=-18---\left(2\right)\\ \text{From}\left(\text{1}\right),p=4.\\ \text{Substitute }p=4\text{ into }\left(2\right):\\ -\frac{{\left(4\right)}^2}{8}+q=-18\\ -\frac{16}{8}+q=-18\\ q=-18+2\\ =-16\end{array}$


(b)
$\begin{array}{l}f\left(x\right)=2x^2-4x-16\\ f\left(x\right)=0\text{ when it cuts }x\text{-axis}\\ 2x^2-4x-16=0\\ x^2-2x-8=0\\ \left(x-4\right)\left(x+2\right)=0\\ x=4,-2\\ \text{Graph }f\left(x\right)\text{ cuts }x\text{-axis at }x=-2\text{ and }x=4.\end{array}$

(c)