What is the vertex form of y= 4x^2-5x-1 ?

Aug 15, 2017

The vertex form is: $y = 4 {\left(x - \frac{5}{8}\right)}^{2} - \frac{41}{16}$.

Refer to the explanation for the process.

Explanation:

$y = 4 {x}^{2} - 5 x - 1$ is a quadratic formula in standard form:

$a {x}^{2} + b x + c$,

where:

$a = 4$, $b = - 5$, and $c = - 1$

$y = a {\left(x - h\right)}^{2} + k$,

where:

$h$ is the axis of symmetry and $\left(h , k\right)$ is the vertex.

The line $x = h$ is the axis of symmetry. Calculate $\left(h\right)$ according to the following formula, using values from the standard form:

$h = \frac{- b}{2 a}$

$h = \frac{- \left(- 5\right)}{2 \cdot 4}$

$h = \frac{5}{8}$

Substitute $k$ for $y$, and insert the value of $h$ for $x$ in the standard form.

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

Simplify.

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

Simplify.

$k = \frac{100}{64} - \frac{25}{8} - 1$

Multiply $- \frac{25}{8}$ and $- 1$ by an equivalent fraction that will make their denominators $64$.

$k = \frac{100}{64} - \frac{25}{8} \left(\frac{8}{8}\right) - 1 \times \frac{64}{64}$

$k = \frac{100}{64} - \frac{200}{64} - \frac{64}{64}$

Combine the numerators over the denominator.

$k = \frac{100 - 200 - 64}{64}$

$k = - \frac{164}{64}$

Reduce the fraction by dividing the numerator and denominator by $4$.

$k = \frac{- 164 \div 4}{64 \div}$

$k = - \frac{41}{16}$

Summary

$h = \frac{5}{8}$

$k = - \frac{41}{16}$

Vertex Form

$y = 4 {\left(x - \frac{5}{8}\right)}^{2} - \frac{41}{16}$

graph{y=4x^2-5x-1 [-10, 10, -5, 5]}