# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function f(x) = -x² + 5x?

Aug 27, 2017

Axis of symmetry: $x = 2.5 \forall y \in \mathbb{R}$
${f}_{\max} = f \left(2.5\right) = 6.25$

#### Explanation:

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

$f \left(x\right)$ is a quadratic function of the form: $a {x}^{2} + b x + c$

Where: $a = - 1 , b = 5 , c = 0$

The graph of $f \left(x\right)$ is a parabola with axis of symmetry where $x = \frac{- b}{2 a}$

in this case

$x = \frac{- 5}{-} 2 = \frac{5}{2} = 2.5$

Hence, the axis of symmetry of $f \left(x\right)$ is the vertical line $x = 2.5 \forall y \in \mathbb{R}$

Since the coefficient of ${x}^{2} < 0$ $f \left(x\right)$ will have a maximum value on the axis of symmetry.

${f}_{\max} = f \left(\frac{5}{2}\right)$

$= - {\left(\frac{5}{2}\right)}^{2} + 5 \cdot \left(\frac{5}{2}\right)$

$= - \left(\frac{25}{4}\right) + \frac{25}{2}$

$= - \frac{25}{4} + \frac{50}{4} = \frac{25}{4} = 6.25$

Hence, ${f}_{\max} = f \left(2.5\right) = 6.25$

Both the axis of symmetry and ${f}_{\max}$ can be seen on the graph of $f \left(x\right)$ below.
graph{(y+x^2-5x)(x-2.5-0.000000001y)=0 [-14.24, 14.23, -7.11, 7.13]}