3.6.1 Fungsi Kuadratik, SPM Praktis (Soalan Pendek)

Soalan 3:
Garis lurus y = 5x – 1 tidak bersilang dengan lengkung y = 2x² + x + h. Carikan julat nilai h.

Penyelesaian:

\begin{array}{l} y = 5 x - 1 ...... (1) \\ y = 2 x^{2} + x + h ...... (2) \\ Gantikan (1) ke dalam (2), \\ 5 x - 1 = 2 x^{2} + x + h \\ 2 x^{2} + x + h - 5 x + 1 = 0 \\ 2 x^{2} - 4 x + h + 1 = 0 \\ b^{2} - 4 a c < 0 \\ {(- 4)}^{2} - 4 (2) (h + 1) < 0 \\ 16 - 8 h - 8 < 0 \\ 8 < 8 h \\ h > 1 \end{array}

Soalan 4:
Cari nilai maksimum bagi fungsi 5 – x – 2x² , dan nilai x apabila ini berlaku.

Penyelesaian:

\begin{array}{l} 5 - x - 2 x^{2} \\ = - 2 x^{2} - x + 5 \\ = - 2 [x^{2} + \frac{1}{2} x - \frac{5}{2}] \\ = - 2 [x^{2} + \frac{1}{2} x + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{1}{16} - \frac{5}{2}] \\ = - 2 [{(x + \frac{1}{4})}^{2} - \frac{41}{16}] \\ = - 2 {(x + \frac{1}{4})}^{2} + 5 \frac{1}{8} \end{array}

\begin{array}{l} Nilai 5 - x - 2 x^{2} adalah maksimum apabila \\ 2 {(x + \frac{1}{4})}^{2} = 0 \\ x = - \frac{1}{4} \\ Nilai maksimum bagi 5 - x - 2 x^{2} ialah 5 \frac{1}{8} . \end{array}