Can the graph of a polynomial function have no y-intercept?

Answer 1

See en.wikipedia.org/wiki/Polynomial

Peter B's example of 1/x doesn't fit the definition of a polynomial (the exponent is negative).

Brad's answer (x=3) is not a polynomial in x. (y=3 would be a polynomial in x (with exponent 0), and it DOES have a y-intercept.)

The answer is no. There must be a y-intercept.

Answer 2

No, each and every polynomial function (in the variable x) would have precisely one y-intercept. (you in basic terms replace x with 0 to locate it.) The area of a polynomial is all genuine numbers. for that reason, 0 is in the area, so which you will constantly be waiting to interchange x with 0 to locate the y-intercept. that's attainable to no longer have an x-intercept. P(x) = x^2 + a million does not have an x-intercept. notice: x = 5 (or the different variety) isn't a polynomial function. additionally, notice that y = a million/x isn't a polynomial. Exponential applications are no longer polynomials (they're exponential applications) and sq. root equations are additionally no longer polynomials.

Answer 3

no - It cannot,

Since a polynomial is a continuous function by definition and that its domain is from negative infinite to positive infinite - It is bound to be passing through x=0 and therefore cross the y-axis (or have a y-intercept).

Note that it might not have a x-intercept. For instance p(x) = x^2+1 is always greater than 1, so will never cross the x-axis.

Answer 4

yes and no, it could be a sideways perabola, but it wouldn't be a function (line test) so. I guess if i had to answer to ur question specifically id say NO.

Answer 5

Yes, it does NOT have to have a y-intercept.

For example: what is x=3? Where's the y-intercept there?

There isn't one.

Answer 6

I think so: if it's undefined (appears as a straight vertical line on the graph)

Answer 7

with a restriction on the domain, of course it can.