How do you find the vertex of #y = 8x + 5x^2 – 3#?

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Answer 1 · 2016-02-11T14:32:51

Vertex #(h, k)=(-4/5, -31/5)#

#y=8x+5x^2-3#

rearrange first

#y=5x^2+8x-3#

Do factor out the 5 to make the coefficient of #x^2# be one.

#y=5(x^2+8/5x)-3#

#y=5(x^2+8/5x+16/25-16/25)-3#

#y=5(x+4/5)^2-16/5-3#

#y=5(x+4/5)^2-31/5#

#y+31/5=5(x+4/5)^2#

#y--31/5=5(x--4/5)^2#

kindly see the graph.
graph{y=8x+5x^2-3[-20,20,-10,10]}

Have a nice day from the Philippines