Graphing f(x)=√x but shifted 2 units to the left and then reflected

[naushaan]

Homework Statement

Hey everyone, one of the questions in my assignment is asking me to graph the shape of f(x)=√x, but shifted 2 units to the left and then reflected in both the x-axis and the y-axis. The question is also asking me to find g(x).

Relevant Equations

f(x)=√x

I drew the graph for f(x)=√x, and then shifted it 2 units to the right. I then drew another graph to reflect it from both x and y axis. Not sure if this is correct though? I'm also quite stuck on trying to write the equation for this final graph.
I think the equation after shifting and reflecting (only on the x axis) would be f(x)=-√(x+2). Not sure how to get the equation for both reflections on the x and y .

 

[Mark44]

Homework Statement:: Hey everyone, one of the questions in my assignment is asking me to graph the shape of f(x)=√x, but shifted 2 units to the left and then reflected in both the x-axis and the y-axis. The question is also asking me to find g(x).
Relevant Equations:: f(x)=√x

View attachment 258081

I drew the graph for f(x)=√x, and then shifted it 2 units to the right. I then drew another graph to reflect it from both x and y axis. Not sure if this is correct though? I'm also quite stuck on trying to write the equation for this final graph. I think for the second graph (before the reflection) the equation would be f(x)=√x -2.

View attachment 258083

You have the basic idea, but your graphs could use some more work. You're working on graph paper, but the graphs don't show this. In the first graph, the curve should go through (0, 0), (1, 1), and (4, 2). In your first graph, 1 is three units up, but 2 is only two more units up.

Every point on the translated graph should be shifted two units left, so it should go through (-2, 0), (-1, 1), and (2, 2). For the next step, a reflection across the y-axis, any point (x, y) on the unreflected graph will now be at (-x, y), and similar for the graph reflected across the x-axis.

 

[naushaan]

thank you for your help! my graphs are not entirely accurate because the question is asking me to write the equation, the graphs are there to help me get to that! I'm struggling with getting to the equation.

 

[epenguin]

I agree with your first two sketches but not the third. Do this in two steps. They say reflected in both axes as though it didn't make any difference what order you do it in. Does it?

For the algebraic expression, for any given y, what x gives you that same y after the translation?

 

[neilparker62]

thank you for your help! my graphs are not entirely accurate because the question is asking me to write the equation, the graphs are there to help me get to that! I'm struggling with getting to the equation.

For the general function f(x), write the transforms for:

a) a translation 2 units left
b) reflection in x-axis
c) reflection in y-axis

Then combine them and apply to the particular f(x) you are working with. As per post #4, does it make any difference if the reflection transforms are applied in different orders?

 

[archaic]

To move a function along the $x$-axis by $a$ units to the left, you consider $f(x+a)$ (you see that you get $f(0)$ when $x=-a$), and if to the right, then you consider $f(x-a)$ (same concept but positive).
To reflect it about the $x$-axis means that whatever was up goes down and inversely, so you consider $-f(x)$.
To reflect it about the $y$-axis means that whatever was right goes left and inversely, so that when you evaluate $f(a)$ you want to get $f(-a)$, hence you consider $f(-x)$.
Now combine these to get an expression for your function and take some points to draw.

Example: I want to shift $f(x)=x^3$ to the right by $4$ units, then reflect it about the $x$-axis, then move it $2.4$ units to the left, then reflect it about the $y$-axis.
1) $f_1(x)=(x-4)^3$
2) $f_2(x)=-f_1(x)$
3) $f_3(x)=f_2(x+2.4)$
4) $f_4(x)=f_3(-x)$
Thus $f_4(x)=f_3(-x)=f_2(-x+2.4)=-f_1(-x+2.4)=-((-x+2.4)-4)^3$

 

[epenguin]

Intake back my disagreement of #4. It is easy to misread the question as apply both transformations but you are only asked to apply them alternately, not successively - maybe.

 

[naushaan]

thank you so much for your help!
following these steps, this is what I've gotten:
1) shifting 2 units to the left: f(x)=√(x+2)
2) reflecting on the x axis: f(x)= -(√(x+2))
3) reflecting on the y axis: f(x)=-(-√(x+2))

Although this brings us back to the equation f(x)=√(x+2) since the negatives cancel out. I'm not sure what I'm doing wrong.

 

[archaic]

thank you so much for your help!
following these steps, this is what I've gotten:
1) shifting 2 units to the left: f(x)=√(x+2)
2) reflecting on the x axis: f(x)= -(√(x+2))
3) reflecting on the y axis: f(x)=-(-√(x+2))

Although this brings us back to the equation f(x)=√(x+2) since the negatives cancel out. I'm not sure what I'm doing wrong.

your last step is wrong.

To reflect it about the $y$-axis means that whatever was right goes left and inversely, so that when you evaluate $f(a)$ you want to get$f(-a)$, hence you consider $f(-x)$.

you have $f(x)=-\sqrt{x+2}$, what is $f(-x)$?

 

[epenguin]

You should be able to work out tests of whether an answer to a problem like this is correct or not - and it will be a great advantage for it to become a habi.

 

[epenguin]

Answer not completed but anyway...

 

[neilparker62]

Cool graphs - what software is that ?

 

[epenguin]

It's called EduCalc - somewhat limited but nice for things like this.