Graph

**symmetry** · January 15th 2007, 12:32 PM

Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.

NOTE:

What is the mirror-like image of the graph y = x?

**AfterShock** · January 15th 2007, 12:45 PM

Originally Posted by symmetry

Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.

NOTE:

What is the mirror-like image of the graph y = x?

y = (sqrt(x))^2 is the same as --> y = x; y = sqrt(x^2) --> y = abs(x)

The mirror image is y = -x

**topsquark** · January 15th 2007, 12:46 PM

Originally Posted by symmetry

Graph y = sqrt{x^2}, y = x, y |x|, and y = (sqrt{x})^2, NOTING which graphs are the SAME.

NOTE:

What is the mirror-like image of the graph y = x?

I'm not going to graph these, you should be able to do that yourself. However I will note that only two of these graphs are the same. The trick is in the last graph: $\sqrt{x^2}$ and $(\sqrt{x})^2$ are "apparently" the same since each reduces to x, but note that in the first expression the domain is $(-\infty, \infty )$ and in the last second expression is $[0, \infty )$ .

$y = \sqrt{x^2}$ and $y = x$ aren't the same either since $\sqrt{.}$ only returns a positive value. It is for this very reason that $y = \sqrt{x^2}$ and $y = |x|$ have the same graph.

Originally Posted by symmetry

What is the mirror-like image of the graph y = x?

Depends on which "mirror" you are talking about. The typical mirror "planes" are the x and y axes, and the lines y = x and y = -x.

y = x reflected over the x-axis is y = -x.
y = x reflected over the y-axis is y = -x.
y = x reflected over the line y = x is y = x.
y = x reflected over the line y = -x is y = x.

-Dan

**topsquark** · January 15th 2007, 12:50 PM

Originally Posted by AfterShock

y = (sqrt(x))^2 is the same as --> y = x; y = sqrt(x^2) --> y = abs(x)

The mirror image is y = -x

I've posted a graph of both functions. $y = \sqrt{x^2}$ is in solid red and $y = x$ is the dotted blue. It's hard to see the overlap in the first quadrant, but you know it's there.

-Dan

**AfterShock** · January 15th 2007, 12:53 PM

Originally Posted by topsquark

I've posted a graph of both functions. $y = \sqrt{x^2}$ is in solid red and $y = x$ is the dotted blue. It's hard to see the overlap in the first quadrant, but you know it's there.

-Dan

Indeed; as far as I could tell, he was asking for the mirror image of just y = x, not the original reflected over y = x.

Wasn't very clear.

**topsquark** · January 15th 2007, 01:50 PM

Originally Posted by AfterShock

Indeed; as far as I could tell, he was asking for the mirror image of just y = x, not the original reflected over y = x.

Wasn't very clear.

Actually I wasn't very clear, sorry. I was responding to your comment that $y = (\sqrt{x})^2$ and $y = x$ have the same graph. But I graphed the wrong function.

These aren't the same because the first only has the domain $[0, \infty )$ , whereas the second has a domain of all real numbers.

-Dan

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