**3.6 Quadratic Functions, SPM Practice (Short Questions)**

**Question 1:**

Find the minimum value of the function

*f*(*x*) = 2*x*^{2 }+ 6*x*+ 5. State the value of*x*that makes*f*(*x*) a minimum value.

*Solution:*By completing the square for

*f*(*x*) in the form of*f*(*x*) =*a*(*x*+*p*)^{2}+*q*to find the minimum value of*f*(*x*).
$\begin{array}{l}f\left(x\right)=2{x}^{2}+6x+5\\ =2\left[{x}^{2}+3x+\frac{5}{2}\right]\\ =2\left[{x}^{2}+3x+{\left(3\times \frac{1}{2}\right)}^{2}-{\left(3\times \frac{1}{2}\right)}^{2}+\frac{5}{2}\right]\end{array}$

$\begin{array}{l}=2\left[{\left(x+\frac{3}{2}\right)}^{2}-\frac{9}{4}+\frac{5}{2}\right]\\ =2\left[{\left(x+\frac{3}{2}\right)}^{2}+\frac{1}{4}\right]\\ =2{\left(x+\frac{3}{2}\right)}^{2}+\frac{1}{2}\end{array}$

Since

Therefore *a*= 2 > 0,*f*(

*x*) has a minimum value when $x=-\frac{3}{2}.$ . The minimum value of

*f*(

*x*) = ½.

**Question 2:**

The quadratic function

*f*(*x*) = –*x*^{2 }+ 4*x*+*k*^{2}, where*k*is a constant, has a maximum value of 8.Find the possible values of

*k*.

*Solution:**f*(

*x*) = –

*x*

^{2 }+ 4

*x*+

*k*

^{2}

*f*(

*x*) = –(

*x*

^{2 }– 4

*x*) +

*k*

^{2}← [completing the square for

*f*(

*x*) in

the form of

*f*(*x*) =*a*(*x*+*p*)^{2}+*q*]*f*(

*x*) = –[

*x*

^{2 }– 4

*x +*(–2)

^{2}– (–2)

^{2}] +

*k*

^{2}

*f*(

*x*) = –[(

*x*

^{ }– 2)

^{2}– 4] +

*k*

^{2}

*f*(

*x*) = –(

*x*

^{ }– 2)

^{2}+ 4 +

*k*

^{2}

Given the maximum value is 8.

Therefore, 4 +

*k*^{2 }= 8*k*

^{2 }= 4

*k*

^{ }= ±2

Good questions helps me to clear my problem