# Thread: What kind of a graph is this?

1. ## What kind of a graph is this?

y= (2x-2x^2)^(1/2)

Hyperbola?

And also-- in a Taylor series approximation, when the questions asks you to use up to the fifth term (NOT the fifth power term), does that just mean the first five numbers in the series?

for example, if the series was 1 + x + x^2 + x^3 + x^4 + x^5 etc. etc., would you plug in x from the first number (1) until the fourth one (x^4) ? Or would you start with x and then go to x^5?

2. Hello, MegaVortex7!

$y \;=\; (2x-2x^2)^{\frac{1}{2}}$ . . .Hyperbola? . no

We have: . $y \;=\;\sqrt{2x-2x^2}$

Square: . $y^2 \;=\;2x - 2x^2 \quad\Rightarrow \quad 2x^2 - 2x + y^2 \:=\:0$

Complete the square: . $2(x^2 - x) + y^2 \:=\:0 \quad\Rightarrow\quad 2\left(x^2 - x + \tfrac{1}{4}\right) + y^2 \:=\:\tfrac{1}{2}$

. . $2\left(x - \tfrac{1}{2}\right)^2 + y^2 \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \frac{(x-\frac{1}{2})^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{2}} \;=\;1$

This is an ellipse; its major axis is vertical.
. . The semimajor axis is: . $a \,=\,\tfrac{\sqrt{2}}{2}$
. . The semiminor axis is: . $b \,=\,\tfrac{1}{2}$

And the graph is the upper half of this ellipse.

3. Ah ok, thanks! And can anyone answer my second question?

Repeated:
And also-- in a Taylor series approximation, when the questions asks you to use up to the fifth term (NOT the fifth power term), does that just mean the first five numbers in the series?

for example, if the series was 1 + x + x^2 + x^3 + x^4 + x^5 etc. etc., would you plug in x from the first number (1) until the fourth one (x^4) ? Or would you start with x and then go to x^5?

4. Originally Posted by Soroban
Hello, MegaVortex7!

We have: . $y \;=\;\sqrt{2x-2x^2}$

Square: . $y^2 \;=\;2x - 2x^2 \quad\Rightarrow \quad 2x^2 - 2x + y^2 \:=\:0$

Complete the square: . $2(x^2 - x) + y^2 \:=\:0 \quad\Rightarrow\quad 2\left(x^2 - x + \tfrac{1}{4}\right) + y^2 \:=\:\tfrac{1}{2}$

. . $2\left(x - \tfrac{1}{2}\right)^2 + y^2 \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \frac{(x-\frac{1}{2})^2}{\frac{1}{4}} + \frac{y^2}{\frac{1}{2}} \;=\;1$

This is an ellipse; its major axis is vertical.
. . The semimajor axis is: . $a \,=\,\tfrac{\sqrt{2}}{2}$
. . The semiminor axis is: . $b \,=\,\tfrac{1}{2}$

And the graph is the upper half of this ellipse.
The graph:

5. Originally Posted by MegaVortex7
Ah ok, thanks! And can anyone answer my second question?

Repeated:
And also-- in a Taylor series approximation, when the questions asks you to use up to the fifth term (NOT the fifth power term), does that just mean the first five numbers in the series?

for example, if the series was 1 + x + x^2 + x^3 + x^4 + x^5 etc. etc., would you plug in x from the first number (1) until the fourth one (x^4) ? Or would you start with x and then go to x^5?
use the first five terms of the series ...

1 + x + x^2 + x^3 + x^4