vértices 3x^2+2x+5y-6=0

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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vertices 3x²+2x+5y−6=0

Solution

Maximum (−13 , 1915 )

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

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3x²+2x+5y−6=0

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

Rewrite 3x²+2x+5y−6=0 in the form y=ax²+bx+c

y=−3x²5 −2x5 +65

The parabola parameters are:

a=−35 , b=−25 , c=65

x_v=−b2a

x_v=−(−25 )2(−35 )

Simplify −−25 2(−35 ) : −13

x_v=−13

Plug in x_v=−13 to find the y_v value

y_v=1915

Therefore the parabola vertex is

(−13 , 1915 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−35

Maximum (−13 , 1915 )

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

vértices 3x^2+2x+5y-6=0

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