# How do you find the axis of symmetry, and the maximum or minimum value of the function 2x^2-5x+4=0?

Jun 18, 2016

Min $\left(\frac{5}{4} , \frac{7}{4}\right)$

#### Explanation:

$y = 2 {x}^{2} - 5 x + 4$
x-coordinate of vertex, or of the axis of symmetry:
$x = - \frac{b}{2 a} = \frac{5}{4}$
y-coordinate of vertex, with $x = \frac{5}{4}$:
$y \left(\frac{5}{4}\right) = 2 \left(\frac{25}{16}\right) - 5 \left(\frac{5}{4}\right) + 4 = \frac{25}{8} - \frac{25}{4} + 4 = \frac{7}{4}$.
Since a > 0, the parabola opens upward, there is min at vertex.
Min $\left(\frac{5}{4} , \frac{7}{4}\right)$
graph{2x^2 - 5x + 4 [-5, 5, -2.5, 2.5]}