Transformation of a function 3.5 #27: f(x)=√x+4

[Illvoices]

Hi can someone tell me how'd the answer for this problem ended up starting in the second quadrant

f(x)=√x+4

 

[Harry_the_cat]

Hi can someone tell me how'd the answer for this problem ended up starting in the second quadrant

f(x)=√x+4

Do you mean \(\displaystyle f(x) = \sqrt{x+4}\)?

And do you mean "why the graph starts in the second quadrant"?

 

[Illvoices]

yes thanks for replying that is exactly what i mean cat

 

[Harry_the_cat]

yes thanks for replying that is exactly what i mean cat

Well think about the values that x can be (ie the domain of the function). x can't be less than -4, do you see why? Ix x=-4, then y=0. So this is your "starting point" for the graph.

 

[Otis]

... that is exactly what i mean ...

Great! But you typed something else.

To show that adding four happens inside the radical sign, we need to type grouping symbols, like this:

f(x) = √(x+4)

From a transformation point-of-view, we're adding four to the input of the square-root function.

Adding a constant to the input of a function has the effect of horizontally shifting the graph by that amount; if the added constant is positive, the graph shifts to the left. If the added constant is negative, the graph shifts to the right.

Therefore, the graph of √(x+4) is the graph of √x shifted 4 units to the left.