Question 13:
It is given that the graph of the quadratic function f(x) = kx2 + 3kx + 10k2, such that k is a constant, has a maximum value of 49/4.
(a) By using the method of completing the square, find the value of k. [5 marks]
(b) Hence, sketch the graph of f(x) for -6 ≤ x ≤ 3. [3 marks]
Answer:
(a)
$$ \begin{aligned} & f(x)=k x^2+3 k x+10 k^2 \\ & f(x)=k\left(x^2+3 x+10 k\right) \\ & f(x)=k\left[x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+10 k\right] \\ & f(x)=k\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+10 k\right] \\ & f(x)=k\left(x+\frac{3}{2}\right)^2-\frac{9}{4} k+10 k^2 \end{aligned} $$
$$ \begin{aligned} &f(x) \text { maximum }, k<0\\ &\therefore k=-1 \end{aligned} $$
$$ \begin{aligned} &\ \text { Maximum } \text { value }=\frac{49}{4}\\ &\begin{aligned} -\frac{9}{4} k+10 k^2 & =\frac{49}{4} \\ -9 k+40 k^2 & =49 \\ 40 k^2-9 k-49 & =0 \\ (40 k-49)(k+1) & =0 \\ k=\frac{49}{40} \quad, \quad k & =-1 \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} &\begin{aligned} f(x) & =(-1)\left(x+\frac{3}{2}\right)^2-\frac{9}{4}(-1)+10(-1)^2 \\ f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ f(-6) & =-\left(-6+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned}\\ &\begin{aligned} f(3) & =-\left(3+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ -\left(x+\frac{3}{2}\right)^2+\frac{49}{4} & =0 \\ -\left(x+\frac{3}{2}\right)^2 & =-\frac{49}{4} \\ \left(x+\frac{3}{2}\right)^2 & =\frac{49}{4} \\ x+\frac{3}{2} & = \pm \frac{7}{2} \end{aligned} $$
$$ \begin{array}{lll} x=-\frac{3}{2}+\frac{7}{2} & , & x=-\frac{3}{2}-\frac{7}{2} \\ x=2 & , & x=-5 \end{array} $$
It is given that the graph of the quadratic function f(x) = kx2 + 3kx + 10k2, such that k is a constant, has a maximum value of 49/4.
(a) By using the method of completing the square, find the value of k. [5 marks]
(b) Hence, sketch the graph of f(x) for -6 ≤ x ≤ 3. [3 marks]
Answer:
(a)
$$ \begin{aligned} & f(x)=k x^2+3 k x+10 k^2 \\ & f(x)=k\left(x^2+3 x+10 k\right) \\ & f(x)=k\left[x^2+3 x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+10 k\right] \\ & f(x)=k\left[\left(x+\frac{3}{2}\right)^2-\frac{9}{4}+10 k\right] \\ & f(x)=k\left(x+\frac{3}{2}\right)^2-\frac{9}{4} k+10 k^2 \end{aligned} $$
$$ \begin{aligned} &f(x) \text { maximum }, k<0\\ &\therefore k=-1 \end{aligned} $$
$$ \begin{aligned} &\ \text { Maximum } \text { value }=\frac{49}{4}\\ &\begin{aligned} -\frac{9}{4} k+10 k^2 & =\frac{49}{4} \\ -9 k+40 k^2 & =49 \\ 40 k^2-9 k-49 & =0 \\ (40 k-49)(k+1) & =0 \\ k=\frac{49}{40} \quad, \quad k & =-1 \end{aligned} \end{aligned} $$
(b)
$$ \begin{aligned} &\begin{aligned} f(x) & =(-1)\left(x+\frac{3}{2}\right)^2-\frac{9}{4}(-1)+10(-1)^2 \\ f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ f(-6) & =-\left(-6+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned}\\ &\begin{aligned} f(3) & =-\left(3+\frac{3}{2}\right)^2+\frac{49}{4} \\ & =-8 \end{aligned} \end{aligned} $$
$$ \begin{aligned} f(x) & =-\left(x+\frac{3}{2}\right)^2+\frac{49}{4} \\ -\left(x+\frac{3}{2}\right)^2+\frac{49}{4} & =0 \\ -\left(x+\frac{3}{2}\right)^2 & =-\frac{49}{4} \\ \left(x+\frac{3}{2}\right)^2 & =\frac{49}{4} \\ x+\frac{3}{2} & = \pm \frac{7}{2} \end{aligned} $$
$$ \begin{array}{lll} x=-\frac{3}{2}+\frac{7}{2} & , & x=-\frac{3}{2}-\frac{7}{2} \\ x=2 & , & x=-5 \end{array} $$