2.13.5 Quadratic Functions, SPM Practice (Paper 2)

Question 9:
Given that the quadratic function f(x) = 2x² – px + p has a minimum value of –18 at x = 1.
(a) Find the values of p and q.
(b) With the value of p and q found in (a), find the values of x, where graph f(x) cuts the x-axis.
(c) Hence, sketch the graph of f(x).

Solution:
(a)
$\begin{array}{l}f\left(x\right)=2x^2-px+q\\ =2\left[x^2-\frac{p}{2}x+\frac{q}{2}\right]\\ =2\left[{\left(x+\frac{-p}{4}\right)}^2-{\left(\frac{-p}{4}\right)}^2+\frac{q}{2}\right]\\ =2\left[{\left(x-\frac{p}{4}\right)}^2-\frac{p^2}{16}+\frac{q}{2}\right]\\ =2{\left(x-\frac{p}{4}\right)}^2-\frac{p^2}{8}+q\end{array}$


$\begin{array}{l}\therefore \frac{p}{4}=1------\left(1\right)\\ \text{and }-\frac{p^2}{8}+q=-18---\left(2\right)\\ \text{From}\left(\text{1}\right),p=4.\\ \text{Substitute }p=4\text{ into }\left(2\right):\\ -\frac{{\left(4\right)}^2}{8}+q=-18\\ -\frac{16}{8}+q=-18\\ q=-18+2\\ =-16\end{array}$


(b)
$\begin{array}{l}f\left(x\right)=2x^2-4x-16\\ f\left(x\right)=0\text{ when it cuts }x\text{-axis}\\ 2x^2-4x-16=0\\ x^2-2x-8=0\\ \left(x-4\right)\left(x+2\right)=0\\ x=4,-2\\ \text{Graph }f\left(x\right)\text{ cuts }x\text{-axis at }x=-2\text{ and }x=4.\end{array}$

(c)

Question 10:
(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.