Quadratic Functions, SPM Practice (Short Questions)

Question 3:
The straight line y = 5x – 1 does not intersect with the curve y = 2x² + x + h.
Find the range of values of h.

Solution:
$\begin{array}{l}y=5x-1\mathrm{......}\text{ (1)}\\ y=2x^2+x+h\mathrm{......}\text{ (2)}\\ \text{Substitute (1) into (2),}\\ 5x-1=2x^2+x+h\\ 2x^2+x+h-5x+1=0\\ 2x^2-4x+h+1=0\\ \\ b^2-4ac<0\\ {\left(-4\right)}^2-4\left(2\right)\left(h+1\right)<0\\ 16-8h-8<0\\ 8<8h\\ h>1\end{array}$

Question 4:
Find the maximum value of the function 5 – x – 2x² , and the corresponding value of x.

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


$\begin{array}{l}5-x-2x^2\text{ has a maximum value when}\\ 2{\left(x+\frac{1}{4}\right)}^2=0\\ x=-\frac{1}{4}\\ \text{The maximum value of }5-x-2x^2\text{ is }5\frac{1}{8}.\end{array}$