# What is the vertex form of  f(x) = -5x^2-2x-3 ?

The vertex form

${\left(x - - \frac{1}{5}\right)}^{2} = - \frac{1}{5} \cdot \left(y - - \frac{14}{5}\right)$

#### Explanation:

From the given $f \left(x\right) = - 5 {x}^{2} - 2 x - 3$, let us use $y$ in place of $f \left(x\right)$ for simplicity and then perform "Completing the square method"

$y = - 5 {x}^{2} - 2 x - 3$

$y = - 5 {x}^{2} - 2 \cdot \left(\frac{- 5}{- 5}\right) \cdot x - 3 \text{ }$this is after inserting $1 = \frac{- 5}{- 5}$

we can factor out the -5 from the first two terms excluding the third term -3

y=-5[(x^2-(2x)/(-5)]-3

$y = - 5 \left({x}^{2} + \frac{2 x}{5}\right) - 3$

Add and subtract the value 1/25 inside the grouping symbol. This is obtained from 2/5. Divide 2/5 by 2 then square it. The result is 1/25. So

$y = - 5 \left({x}^{2} + \frac{2 x}{5} + \frac{1}{25} - \frac{1}{25}\right) - 3$

now regroup so that there is a Perfect Square Trinomial
$\left({x}^{2} + \frac{2 x}{5} + \frac{1}{25}\right)$

$y = - 5 \left({x}^{2} + \frac{2 x}{5} + \frac{1}{25}\right) - \left(- 5\right) \left(\frac{1}{25}\right) - 3$

$y = - 5 {\left(x + \frac{1}{5}\right)}^{2} + \frac{1}{5} - 3$

simplify

$y = - 5 {\left(x + \frac{1}{5}\right)}^{2} - \frac{14}{5}$

$y + \frac{14}{5} = - 5 {\left(x + \frac{1}{5}\right)}^{2}$

The vertex form

${\left(x - - \frac{1}{5}\right)}^{2} = - \frac{1}{5} \cdot \left(y - - \frac{14}{5}\right)$

graph{y=-5x^2-2x-3[-10,10,-10,5]}

God bless...I hope the explanation is useful