# How do you find the maximum or minimum of quadratic functions?

Mar 7, 2018

#### Explanation:

For a quadratic function $y = a {x}^{2} + b x + c$, a maximum is there if $a < 0$ and it has a minimum, if $a > 0$. Please see below for details.

We can write $y = a {x}^{2} + b x + c$ as

$y = a \left({x}^{2} + \frac{b}{a} x\right) + c$

= $a \left({x}^{2} + 2 \times \frac{b}{2 a} \times x + {\left(\frac{b}{2 a}\right)}^{2} - {\left(\frac{b}{2 a}\right)}^{2}\right) + c$

= $a \left({x}^{2} + 2 \times \frac{b}{2 a} \times x + {\left(\frac{b}{2 a}\right)}^{2}\right) - a {\left(\frac{b}{2 a}\right)}^{2} + c$

= $a {\left(x + \frac{b}{2 a}\right)}^{2} - {b}^{2} / \left(4 a\right) + c$

= $a {\left(x + \frac{b}{2 a}\right)}^{2} - \frac{{b}^{2} - 4 a c}{4 a}$

Observe that as ${\left(x + \frac{b}{2 a}\right)}^{2}$ is always greater than $0$,

if $a$ is positive, we will have a minima for $y$, when $x + \frac{b}{2 a} = 0$ i.e. $x = - \frac{b}{2 a}$, which will be at $- \frac{{b}^{2} - 4 a c}{4 a}$, and

if $a$ is negative, we will have a maxima for $y$, when $x + \frac{b}{2 a} = 0$ i.e. $x = - \frac{b}{2 a}$, which will be at $- \frac{{b}^{2} - 4 a c}{4 a}$.

Hence to find a maxima or minima for a quadratic function, observe the sign of $a$ and convert the equation, as above, in form $a {\left(x - h\right)}^{2} + k$. Then the corresponding maxima or minima will be $k$, when $x = h$.