Question

The minimum and maximum values of \(f(x) = {x}^{2}+ 4x +17\) are

Answer 1

\(f(x) = {x}^{2} + 4x + 17\)

$a = 1, b = 4, c = 17$

As $a > 0$, minimum value \(=\dfrac{-D}{4a}\)

\(= \dfrac{4ac -{b}^{2}}{4a}\)

\(= c - \dfrac{{b}^{2}}{4a}\)

$= 17 - 4 = 13$

Graph of $f(x)$ will be upward parabola with minimum value $13$

So, maximum value is \(\infty \)