# Can someone help figuring this graphing question?Graph the following function using transformations. Start with the basic function and graph all the stages on a single coordinate plane. f(x)=...

Can someone help figuring this graphing question?

Graph the following function using transformations. Start with the basic function and graph all the stages on a single coordinate plane.

f(x)= 1/6(x-6)^4

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*Graph `f(x)=1/6(x-6)^4` using transformations.*

Given `f(x)=AfB(x-4)+k` we observe the following:

(a) "A" performs a vertical stretch or compression on f(x). Also, if A<0 it reflects f(x) across the x-axis.

(b) "B" performs a horizontal stretch/compression on f(x). If B<0 it reflects across the y-axis. ** Usually we can ignore B except in the case of periodic functions like sin or cos, where it affects the period.

(c) "h" is a horizontal translation

(d) "k" is a vertical tranlation

So:

(1) Identify the basic function you are transforming. Here the function is `f(x)=x^4`

(2) h=6 so move the graph right 6 units

(3) A=1/6; this is a vertical compression. Each y-value is multiplied by 1/6.(e.g. if `x^4=6` , then `1/6x^4=1` )

So, putting it all together on one coordinate system:

original f in black, with horizontal translation in blue; vertical compression after translation in red:

You need to determine what is the restriction of the domain of definition of the function. Since the denominator must not be zero => x != 6. => the domain of definition of function is R-{6}.

The graph of function has vertical asymptotes at x=6.

You should also check if the graph has horizontal asymptotes.

lim_(x->oo) 1/6(x-6)^4 = lim_(x->-oo) 1/6(x-6)^4 = 0 => the graph has horizontal asymptotes at y = 0.

You should check if the function has maximal or minimal points. You need to calculate the first derivative.

Use quotient rule => f'(x) = -24(x-6)^3/36(x-6)^8 = -2/3(x-6)^5

Since f'(x) != 0 => the function has no extremes.

The second derivative is not cancelling and the function has no inflection points.