3.9.4 Quadratic Functions, SPM Practice (Long Question)

Question 6:
Quadratic function f(x) = x² – 4px + 5p² + 1 has a minimum value of m² + 2p, where m and p are constants.
(a) By using the method of completing the square, shows that m = p – 1.
(b) Hence, find the values of p and of m if the graph of the quadratic function is symmetry at x = m² – 1.

Solution:
(a)
$\begin{array}{l}f\left(x\right)=x^2-4px+5p^2+1\\ =x^2-4px+{\left(\frac{-4p}{2}\right)}^2-{\left(\frac{-4p}{2}\right)}^2+5p^2+1\\ ={\left(x-2p\right)}^2+p^2+1\\ \\ \text{Minimum value,}\\ m^2+2p=p^2+1\\ m^2=p^2-2p+1\\ m^2={\left(p-1\right)}^2\\ m=p-1\end{array}$

(b)
$\begin{array}{l}x=m^2-1\\ 2p=m^2-1\\ p=\frac{m^2-1}{2}\\ \text{Given }m=p-1\Rightarrow p=m+1\\ m+1=\frac{m^2-1}{2}\\ 2m+2=m^2-1\\ m^2-2m-3=0\\ \left(m-3\right)\left(m+1\right)=0\\ m=3\text{ or }-1\\ \\ \text{When }m=3,\\ p=\frac{3^2-1}{2}=4\\ \\ \text{When }m=-1,\\ p=\frac{{\left(-1\right)}^2-1}{2}=0\end{array}$