# How do you find the axis of symmetry, and the maximum or minimum value of the function y = 4x^2 + 5x – 1?

Jun 22, 2018

vertex: (-5/8, -91/16); " axis of symmetry": x = -5/8
minimum

#### Explanation:

Given: $y = 4 {x}^{2} + 5 x - 1$

When the equation is in standard form: $f \left(x\right) = A {x}^{2} + B x + C$, you can find the vertex and the axis of symmetry as follows:

vertex: $\left(- \frac{B}{2 A} , f \left(- \frac{B}{2 A}\right)\right)$, axis of symmetry: $x = - \frac{B}{2 A}$

$- \frac{B}{2 A} = - \frac{5}{2 \cdot 4} = - \frac{5}{8}$

$f \left(- \frac{5}{8}\right) = 4 {\left(- \frac{5}{8}\right)}^{2} + 5 \left(- \frac{5}{8}\right) - 1$

$= 4 \cdot \left(\frac{25}{64}\right) - \frac{25}{8} - 1$

$= \frac{25}{16} - \frac{50}{16} - \frac{16}{16} = - \frac{91}{16}$

vertex: (-5/8, -91/16); " axis of symmetry": x = -5/8

The vertex will be a minimum because $A$ is positive.