2 Answers | Add Yours
First we will have to isolate y on the left side.
`==gt 7x^2 + 3y^2 = 1`
` ==gt 3y^2 = 1- 7x^2 `
`==gt y^2 = (1-7x^2)/3 `
`==gt y= +- (1-7x^2)/3`
`` Now we notice that we have two possible values to y for each value of x.
Therefore, we have two images for the value of x ( positive value and negative value.
For example, if x= 1 ==> y= 2 and y= -2)
Then, y is NOT a function of x.
Or, we can graph y and test the vertical line.
Isolate y so we can find out if this is indeed a function or not:
`3y^2 = 1 - 7x^2`
`y^2 = ( 1 - 7x^2 ) / 3`
`y = sqrt ( (1 - 7x^2) / 3 )`
We know that the square root means both positive and negative. So when we plug this into a graphing utility, we need to account for both positive and negative. And when you do this, you'll see that the resulting image is that of a circle which does not pass the vertical line test and so is not a function.
We’ve answered 315,619 questions. We can answer yours, too.Ask a question