intercepts of 1/(x^2)

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vertices (y−2)=3(x−5)²

Solution

Minimum (5, 2)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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(y−2)=3(x−5)²

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 (y−2)=3(x−5)² in the form y=ax²+bx+c

y=3x²−30x+77

The parabola parameters are:

a=3, b=−30, c=77

x_v=−b2a

x_v=−(−30)2· 3

Simplify

x_v=5

Plug in x_v=5 to find the y_v value

y_v=2

Therefore the parabola vertex is

(5, 2)

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=3

Minimum (5, 2)

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

intercepts of 1/(x^2)

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