limit as t approaches infinity of (sqrt(t)+t^2)/(6t-t^2)

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vertices y=(2x−1)(5x−6)

Solution

Minimum (1720 , −4940 )

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Find vertex using averaging the zeros

Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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y=(2x−1)(5x−6)

Parabola equation in factored form

The vertex of an up-down facing parabola of the form y=a(x−m)(x−n) is the average of the zeros x_v=m+n2

y=(2x−1)(5x−6)

The parabola parameters are:

a=10, m=12 , n=65

x_v=m+n2

x_v=(12 )+(65 )2

Simplify (12 )+(65 )2 : 1720

x_v=1720

Plug in x_v=1720 to find the y_v value

y_v=−4940

Therefore the parabola vertex is

(1720 , −4940 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=10

Minimum (1720 , −4940 )

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

limit as t approaches infinity of (sqrt(t)+t^2)/(6t-t^2)

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