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

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Answer 1 · 2016-02-27T08:43:30

The vertex form

$(x--1/5)^2=-1/5*(y--14/5)$

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

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

$y=-5x^2-2*((-5)/(-5))*x-3" "$ this is after inserting $1=(-5)/(-5)$

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

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

$y=-5(x^2+(2x)/5)-3$

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

$y=-5(x^2+(2x)/5+1/25-1/25)-3$

now regroup so that there is a Perfect Square Trinomial
$(x^2+(2x)/5+1/25)$

$y=-5(x^2+(2x)/5+1/25)-(-5)(1/25)-3$

$y=-5(x+1/5)^2+1/5-3$

simplify

$y=-5(x+1/5)^2-14/5$

$y+14/5=-5(x+1/5)^2$

The vertex form

$(x--1/5)^2=-1/5*(y--14/5)$

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

God bless...I hope the explanation is useful

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