# St. the base fun. that corr. to the tran. fun.y = –2(–1/3 (x + 2))^2-4 St. the para. and de. the corr. trans.St. the domain and range of the graph of the tran. fun.fun.="function"...

St. the base fun. that corr. to the tran. fun.y = –2(–1/3 (x + 2))^2-4 St. the para. and de. the corr. trans.St. the domain and range of the graph of the tran. fun.

fun.="function"

tran.="transformed"

trans.="transformations"

corr.="corresponding"

para.="parameters"

St.="State"

corr.="corresponds"

de.="describe"

embizze | Certified Educator

Given `f(x)=-2(-1/3(x+2))^2-4` , determine the base function and the transformations applied.

Note: In general, given a base function f(x), then Af(B(x-h))+k has the following properties:

A performs a vertical dilation/compression. Also, if A<0 it reflects f(x) across the x-axis.

B performs a horizontal dilation/compression. (In the case of a noneven function, B<0 reflects across the vertical axis)

h is a horizontal translation.

k is a vertical translation.

(1) The base function is `f(x)=x^2`

(2) `f(x)=(x+2)^2` translates the base function 2 units to the left. (h=-2)

(3) `f(x)=[-1/3(x+2)]^2` dilates the function by a factor of 3 horizontally. (B=-1/3)

(4) `f(x)=-2[-1/3(x+2)]^2` reflects the transformed function across the x-axis, and dilates by a factor of 2 vertically. (A=-2)

(5) `f(x)=-2[-1/3(x+2)]^2-4` translates the transformed equation 4 units down. (k=-4)

For any quadratic function the domain is all real numbers. For any transformed quadratic;B,and h do not affect the range. A determines if there is a maximum or a minimum, and k determines the value of the extrema.

Since this function opens down (A<0) there is a maximum, which is located at (-2,-4) (The function was translated 2 units left and 4 units down.) Thus the range is `y<= -4` .

f(x) in black;f(x+2) in green; f(-1/3(x+2)) in blue; -2f(-1/3(x+2)) in purple; -2f(-1/3(x+2))-4 in red