inverse of sqrt(x+3)

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

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(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

lhopital squeeze chain factoring sandwich

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inverse 1x³

Solution

³√1x

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Solve by:

Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

One step at a time

y=x²−6x+8

Parabola equation in polynomial form

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

The parabola parameters are:

a=1, b=−6, c=8

x_v=−b2a

x_v=−(−6)2· 1

Simplify

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=1

Minimum (3, −1)

Solve instead

domain 1x³

range 1x³

intercepts 1x³

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