Question

how would you graph (x+2)^2/9+(y-3)^2/16=1?

Answer 1

idk how would u?:(

Answer 2

wait

Answer 3

You could use a lot of x,y pairs or, better yet, get y=f(x) by solving the equation and plot that. Move the term in x over to the right-hand side, then take square root.

Answer 4

What you have here is the equation of an ellipse. The general form of an ellipse is
\[\frac{(x-c)^2}{a^2}+\frac{(y-d)^2}{b^2}=1\]
where \((c,d)\) is the center of the ellipse and \(a\) and \(b\) are the lengths of the horizontal and vertical axes, respectively. The larger of the two would be called the major axis, the smaller would be the minor axis.

So you are given \(\dfrac{(x+2)^2}{9}+\dfrac{(y-3)^2}{16}=1\), which means the center is \((-2,3)\), \(a=3\), and \(b=4\). Since \(b>a\), the vertical axis is the major axis. You can see this pretty clearly in the image posted earlier.

If you haven't seen the general form before, you can always do as @douglaswinslowcooper mentioned. This would involve solving for a particular variable in terms of the other, then substituting in values of the independent variable and solving for the other.

For example, you could solve for \(y\) and get
\[y=\sqrt{16\left(1-\dfrac{(x+2)^2}{9}\right)}+3\]
then you would plug in a value of \(x\), say \(x=0\), then solve for \(y\):
\[y=\sqrt{16\left(1-\dfrac{(0+2)^2}{9}\right)}+3~~\Rightarrow~~y=\frac{4\sqrt{5}}{3}+3\approx 5.981\]
So one of the many points on the ellipse is \((0, 5,981.)\). Repeat this process, preferably with values of \(x\) that make the solving for \(y\) manageable.

Answer 5

thank you everyone!!