substitution 1/2 x+1/7 y=4, 1/6 x-2/7 y=-1

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

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□□

log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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vertices (x+3)²=−20(y−1)

Solution

Maximum (−3, 1)

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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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(x+3)²=−20(y−1)

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 (x+3)²=−20(y−1) in the form y=ax²+bx+c

y=−x²20 −3x10 −920 +1

The parabola parameters are:

a=−120 , b=−310 , c=−920 +1

x_v=−b2a

x_v=−(−310 )2(−120 )

Simplify −−310 2(−120 ) : −3

x_v=−3

Plug in x_v=−3 to find the y_v value

y_v=1

Therefore the parabola vertex is

(−3, 1)

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−120

Maximum (−3, 1)

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Solve by: Find vertex using parabola form

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Solve by: Find vertex using averaging the zeros

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

substitution 1/2 x+1/7 y=4, 1/6 x-2/7 y=-1

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