Question

Determine if the following represents a function: x2 + y2 = 16

Answer 1

Take some points. Plot a graph. And take a Vertical Line Test. If at any give point, the vertical line cuts the curve at more than one point, then the given equation is not a function.

Answer 2

if it represents a circle then it does not represent a function right?

Answer 3

Yes correct. That's because for any one single value of x, there are two different values of y. :)
You got it right :)

Answer 4

thanks:) but can you explain how i she figured out it was a circle? and how exactly can you graph it?

Answer 5

@jhannybean usually the circle formula is written \((x-h)^2 + (y-k)^2 = r^2\) where the center is at \((h,k)\) and the radius is \(r\). Your version has the center at \((-h,-k)\)

@jannakay she figured out it was a circle because it matched the pattern of the formula of a circle. to graph it, find the center (from the formula), set a compass to the radius (also from the formula), and mark off a circle around the center. Or plot points that satisfy the equation, but the compass thing is a lot easier!

Answer 6

there's something called the vertical line test: if you can draw a vertical line anywhere on the graph and have it cross the curve twice, it's not a function, because at that value of x, there are two or more values of y.

Answer 7

Either you can put up the points and then plot a graph or use the method or use the above formula.

Answer 8

@whpalmer4 well i know how to use the vertical line test as long as the graph is given. but i don't really know how to make one with just an equation.

Answer 9

Well, after you had graphed it, you could apply the test.

Answer 10

@Utk*4 yeah but how do you know what points to use?

Answer 11

Pick a value of x. Plug it in equation. Solve for y. Repeat with another value for x :-)

Answer 12

so just randon numbers?

Answer 13

to avoid pointless work, you'll want to pick values of x and y that are no more than r away from the center... if your center is at (2,3) and the radius is 1, there's no point in trying x = 5 because that's not going to be part of the circle... if you have the center at (h,k) and radius r (all of which you read off from the formula), I would just plot (h,k+r), (h+r,k), (h,k-r), and (h-r,k) which represent the "compass points" N E S W of the circle, then draw the circle through them.

Answer 14

you know what circle looks like, so once you have those 4 points its not too hard to draw a nice freehand circle through them with a little bit of practice.

Answer 15

Yeah it does make sense. Thank you.. :)

Answer 16

Take Y^2 to R.H.S.
Now substitute any real value of Y in RHS for which you'll get a value of X in LHS. Keep on doing this using different values. After few points, you'll be able to recognize the graph taking shape of a circle. You don't have to draw the circle with accuracy. Just an idea that it would be a circle will solve our problem of it being not a function.

Answer 17

@Utk*4 Thank you. :)

Answer 18

You're welcome. :)