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