Equation of the axis of symmetry of the graph is $x=\frac{2}{3}.$   

Question 5:
Find the range of values of k if the quadratic equation 3(x² – kx – 1) = k – k² has two real and distinc roots.

Solution:
$\begin{array}{l}3\left(x^2-kx-1\right)=k-k^2\\ 3x^2-3kx-3-k+k^2=0\\ 3x^2-3kx+k^2-k-3=0\\ a=3,b=-3k,c=k^2-k-3\\ \\ \text{In cases of two real and distinc roots,}\\ b^2-4ac>0\text{ is applied}\text{.}\\ {\left(-3k\right)}^2-4\left(3\right)\left(k^2-k-3\right)>0\\ 9k^2-12k^2+12k+36>0\\ -3k^2+12k+36>0\\ -k^2+4k+12>0\\ k^2-4k-12<0\\ \left(k+2\right)\left(k-6\right)<0\\ k=-2,6\end{array}$



$\text{The range of values of }k\text{ is }-2<k<6.$

Question 6:
Given that the quadratic equation hx² – (h + 2)x – (h – 4) = 0 has real and distinc roots.  Find the range of values of h.

Solution:
$\begin{array}{l}\text{The quadratic equation }h{x}^2-\left(h+2\right)x-\left(h-4\right)=0\\ \text{has real and distinc roots}\text{.}\\ b^2-4ac>0\text{ is applied}\text{.}\\ {\left(-h-2\right)}^2-4\left(h\right)\left(-h+4\right)>0\\ h^2+4h+4+4h^2-16h>0\\ 5h^2-12h+4>0\\ \left(5h-2\right)\left(h-2\right)>0\\ \\ \text{The coefficient of }{h}^2\text{ is positive, }\\ \text{the region above the }x\text{-axis should be shaded}\text{.}\\ \left(5h-2\right)\left(h-2\right)=0\\ h=\frac{2}{5},2\end{array}$



$\begin{array}{l}\text{The range of values of }h\text{ for }\left(5h-2\right)\left(h-2\right)\text{>0 is}\\ h<\frac{2}{5}\text{ or }h>2.\end{array}$

Example 2 (Combination of Straight Line and a Curve)
The straight line $y=2k+1$   intersects the curve $y=x+\frac{k^2}{x}$   at two distinct points.  Find the range of values of k.

Example 3 (Straight Line does not intersect the curve)
Find the range of values of m for which the straight line $y=mx+6$   does not meet the curve $2x^2-xy=3$  .

Example 1
Find the value of p for which $8y=x+2p$   is a tangent to the curve $2y^2=x+p$  .

Nature of the Roots (Combination of Straight Line and the Curve)
When you have a straight line and a curve, you can solve the equation of the straight line and the curve simultaneously and form a quadratic equation, ax² +bx + c = 0. The discriminant, $b^2-4ac$  gives information about the number of points of intersection.

Solving Quadratic Inequality

Solving Quadratic Inequality - Graph Method

Example 1
Express $y=5+4x-x^2$   in the form $y=a-{(x+b)}^2$   , where a, b, and c  are constants. Hence state the maximum  value of y and the value of x at which it occurs. Sketch the curve $y=5+4x-x^2$   .

Example
Sketch the curve of the quadratic function $f(x)=x^2-<!--\; -\; -->x-<!--\; -\; -->12$

Answer:
The shape of the graph
Since the coefficient of x² is positive, hence the graph is a U shape parabola with a minimum point.

The minimum point of the graph
By completing the square
$\begin{array}{l}f(x)=x^2-x-12\\ f(x)=x^2-x+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2-12\\ f(x)={\left(x-\frac{1}{2}\right)}^2-\frac{1}{4}-12\\ f(x)={\left(x-\frac{1}{2}\right)}^2-12\frac{1}{4}\\ \\ \text{Minimum point = }(\frac{1}{2},-12\frac{1}{4})\end{array}$

For y-intercept, x = 0
$f(0)=(0{)}^2-<!--\; -\; -->(0)-<!--\; -\; -->12=-<!--\; -\; -->12$

For x-intercept, f(x) = 0
$\begin{array}{l}f(x)=x^2-x-12\\ 0=x^2-x-12\\ (x+5)(x-6)=0\\ x=-5\text{ or }x=6\end{array}$

Example 1 
Find the maximum or minimum value of each of the following quadratic function by completing the squares.  In each case, state the value of x at which the function is maximum or minimum.  And also, state the maximum or minimum point and axis of symmetry for each case.

(a) $f(x)=x^2+6x+7$
(b) $f(x)=2x^2-6x+7$
(c) $f(x)=5-2x-x^2$
(d) $f(x)=4+12x-3x^2$

Steps to convert general form of Quadratic Function into completing the square form

General form of quadratic function : $f(x)=a{x}^2+bx+c$

Completing the square form : $f(x)=a{(x+p)}^2+q$

Step 1 : Make sure the coefficient $x^2$ is 1, if not factorize.
Step 2 : Insert $+{\left(\frac{coefficient\text{}\text{}-of\text{}x}{2}\right)}^2-{\left(\frac{coefficient\text{}\text{}-of\text{}x}{2}\right)}^2$
Step 3 : Completing the square [convert $f(x)=a{x}^2+bx+c$ into $f(x)=a{(x+p)}^2+q$ .]