# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function f(x)=x^2-5x+6?

Nov 18, 2017

axis symmetry $X = \frac{5}{2}$

${f}_{\min} \left(\frac{5}{2}\right) = - \frac{1}{4}$

graph below

#### Explanation:

firstly we complete the square

$f \left(x\right) = {x}^{2} - 5 x + 6$

$f \left(x\right) = {x}^{2} - 5 x + \textcolor{red}{{\left(\frac{5}{2}\right)}^{2}} - {\left(\frac{5}{2}\right)}^{2} + 6$

$f \left(x\right) = {\left(x - \frac{5}{2}\right)}^{2} - \frac{25}{4} + 6$

$f \left(x\right) = {\left(x - \frac{5}{2}\right)}^{2} - \frac{1}{4}$

axis of symetry is when bracket$= 0$

$X = \frac{5}{2}$

now ${\left(x - \frac{5}{2}\right)}^{2} \ge 0 \text{ } \forall x \in \mathbb{R}$

$\therefore f \left(x\right) \ge - \frac{1}{4}$

min at $f \left(\frac{5}{2}\right) = - \frac{1}{4}$ graph{x^2-5x+6 [-10, 10, -5, 5]}