2.13.3 Quadratic Functions, SPM Practice (Paper 2 Question 11 – 15)

Question 11:


Diagram above shows the graphs of the curves y = x² + x – kx + 5 and y = 2(x – 3) – 4h that intersect the x-axis at two points. Find
(a) the value of k and of h,
(b) the minimum value of each curve.


Solution:
(a)
$\begin{array}{l}y=x^2+x-kx+5\\ =x^2+\left(1-k\right)x+5\\ ={\left[x+\frac{\left(1-k\right)}{2}\right]}^2-{\left(\frac{1-k}{2}\right)}^2+5\\ \text{axis of symmetry of the graph is}\\ x=-\frac{\left(1-k\right)}{2}\end{array}$

$\begin{array}{l}y=2{\left(x-3\right)}^2-4h\\ \text{axis of symmetry of the graph is }x=3.\\ \therefore -\frac{1-k}{2}=3\\ -1+k=6\\ k=7\end{array}$

$\begin{array}{l}\text{Substitute }k=7\text{ into equation}\\ y=x^2+x-7x+5\\ =x^2-6x+5\\ \text{At }x\text{-axis},y=0;\\ x^2-6x+5=0\\ \left(x-1\right)\left(x-5\right)=0\\ x=1,5\end{array}$

$\begin{array}{l}\text{At point }\left(1,0\right)\\ \text{Substitute }x=1,y=0\text{ into the graph:}\\ y=2{\left(x-3\right)}^2-4h\\ 0=2{\left(1-3\right)}^2-4h\\ 4h=2\left(4\right)\\ 4h=8\\ h=2\end{array}$

(b)
$\begin{array}{l}\text{For }y=x^2-6x+5\\ ={\left(x-3\right)}^2-9+5\\ ={\left(x-3\right)}^2-4\\ \therefore \text{ Minimum value is }-4.\\ \\ \text{For }y=2{\left(x-3\right)}^2-8,\text{ minimum value is}-8.\end{array}$

Question 12:
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 13 (SPM 2009):
The quadratic equation x² – 5x + 6 = 0 has roots h and k, where h > k.
(a) Find
(i) the value of h and of k.
(ii) the range of x if x² – 5x + 6 > 0.
[5 marks]

(b) Using the values of h and k from 2(a)(i), form the quadratic equation which has roots h + 2 and 3k – 2.
[2 marks]


Answer:
(a)(i) 
$$ \begin{aligned} &\begin{array}{r} x^2-5 x+6=0 \\ (x-2)(x-3)=0 \\ x=2 \text { or } x=3 \end{array}\\ &\text { Since } h>k, h=3 \text { and } k=2 \text {. } \end{aligned} $$

(a)(ii)
$$ \begin{array}{r} x^2-5 x+6>0 \\ (x-2)(x-3)>0 \end{array} $$ The range is x < 2 or x > 3.



(b)
$$ \begin{aligned} &\begin{aligned} & h+2=3+2=5 \\ & 3 k-2=3(2)-2=4 \end{aligned}\\ &\text { The quadratic equation is }\\ &\begin{array}{r} x^2-(5+4) x+5(4)=0 \\ x^2-9 x+20=0 \end{array} \end{aligned} $$