3.7.3 Fungsi Kuadratik, SPM Praktis (Soalan Panjang)

Soalan 5:
Diberi fungsi kuadratik f(x) = 2x² – px + p mempunyai nilai minimum –18 pada nilai x = 1.

1. Cari nilai p dan nilai q.
2. Dengan nilai p dan nilai q yang diperoleh, cari nilai-nilai x di mana graf f(x), memotong paksi-x.
3. Seterusnya, lakarkan graf bagi f(x).

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

$\begin{array}{l}\text{Maka},\frac{p}{4}=1------\left(1\right)\\ \text{dan }-\frac{p^2}{8}+q=-18---\left(2\right)\\ \text{Dari}\left(\text{1}\right),p=4.\\ \text{Ganti }p=4\text{ ke dalam }\left(2\right):\\ -\frac{{\left(4\right)}^2}{8}+q=-18\\ -\frac{16}{8}+q=-18\\ q=-18+2\\ =-16\end{array}$


(b)
$\begin{array}{l}f\left(x\right)=2x^2-4x-16\\ \text{Memotong paksi-}x,f\left(x\right)=0.\\ 2x^2-4x-16=0\\ x^2-2x-8=0\\ \left(x-4\right)\left(x+2\right)=0\\ x=4,-2\\ \text{Graf }f\left(x\right)\text{ memotong paksi-}x\text{ di }x=-2\text{ dan }x=4.\end{array}$

(c)