Question 7:
The diagram below shows the graph of the quadratic function f(x) = (x + 3)² + 2h – 6, where h is a constant.


(a) State the equation of the axis of symmetry of the curve.
(b) Given the minimum value of the function is 4, find the value of h.

Solution:
(a)
When x + 3 = 0
 x = –3
Therefore, equation of the axis of symmetry of the curve is x = –3.

(b)
When x + 3 = 0, f(x) = 2h – 6
Minimum value of f(x) is 2h – 6.
2h – 6 = 4
2h = 10
h = 5

Question 8 (4 marks):
The quadratic function f is defined by f(x) = x² + 4x + h, where h is a constant.
(a) Express f(x) in the form (x + m)² + n, where m and n are constants.

(b) Given the minimum value of f(x) is 8, find the value of h.

Solution:
(a)
f(x) = x² + 4x + h
  = x² + 4x + (2)² – (2)² + h
  = (x + 2)² – 4 + h

(b)
Given the minimum value of f(x) = 8
– 4 + h = 8
h = 12

Question 9 (3 marks):
Find the range of values of x such that the quadratic function f(x) = 6 + 5x – x² is negative.

Solution:
(a)
f(x) < 0
6 + 5x – x² < 0
(6 – x)(x + 1) < 0
x < –1, x > 6

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

Question 4:
(a) Find the range of values of k if the equation x² – kx + 3k – 5 = 0 does not have real roots.
(b) Show that the quadratic equation hx² – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Solution:
(a)
$\begin{array}{l}{x}^2-kx+\left(3k-5\right)=0\\ \text{If the above equation has no real root,}\\ \therefore b^2-4ac<0.\\ k^2-4\left(3k-5\right)<0\\ k^2-12k+20<0\\ \left(k-2\right)\left(k-10\right)<0\end{array}$

Graph function y = (k – 2)(k – 10) cuts the horizontal line at k = 2 and k = 10 when b² – 4ac < 0.



The range of values of k that satisfy the inequality above is 2 < k < 10.

(b)
$\begin{array}{l}h{x}^2-\left(h+3\right)x+1=0\\ b^2-4ac={\left(h+3\right)}^2-4\left(h\right)\left(1\right)\\ =h^2+6h+9-4h\\ =h^2+2h+9\\ ={\left(h+\frac{2}{2}\right)}^2-{\left(\frac{2}{2}\right)}^2+9\\ ={\left(h+1\right)}^2-1+9\\ ={\left(h+1\right)}^2+8\end{array}$

The minimum value of (h + 1) + 8 is 8, a positive value. Therefore, b² – 4ac > 0 for all values of h.
Hence, quadratic equation hx² – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.