# What is the axis of symmetry and vertex for the graph y=4x^2+5x-1?

Jul 5, 2017

x_("vertex")=" axis of symmetry"=-5/8

Vertex$\to \left(x , y\right) = \left(- \frac{5}{8} , - \frac{41}{16}\right)$

#### Explanation:

The coefficient of ${x}^{2}$ is positive so the graph is of form $\cup$. Thus the vertex is a minimum.

$y = 4 {x}^{2} + 5 x - 1 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$

$\textcolor{g r e e n}{\underline{\text{Part}}}$ of the process of completing the square gives you:

$y = 4 \left({x}^{2} + \frac{5}{4} x\right) - 1 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(2\right)$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(+ \frac{5}{4}\right) = - \frac{5}{8}$

Substitute for $x \text{ in } E q u a t i o n \left(1\right)$ giving:

${y}_{\text{vertex}} = 4 {\left(- \frac{5}{8}\right)}^{2} + 5 \left(- \frac{5}{8}\right) - 1$

${y}_{\text{vertex}} = - 2 \frac{9}{16} \to - \frac{41}{16}$

Vertex$\to \left(x , y\right) = \left(- \frac{5}{8} , - \frac{41}{16}\right)$