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

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