You could argue that if you meant to write y-1 or y+1 . But it would be rather pedantic. Literally, " y 1 " is meaningless.
YES! A polynomial is made up of terms that are only added, subtracted or multiplied. Division is not allowed. It may have constants, exponents or variables. A monomial is one type of polynomial. y+2y is the same as 4y which is a monomial. It has degree 1 since the exponent on y is 1.
For a polynomial of the form y = p(x) (i.e., some polynomial function of x), having a y-intercept simply means that the polynomial is defined for x = 0 - and a polynomial is defined for any value of "x". As for the x-intercept: from left to right, a polynomial of even degree may come down, not quite reach zero, and then go back up again. A simple example is y = x2 + 1. Why is the situation for "x" and for "y" different? Well, the original equation is a polynomial in "x"; but if you solve for "x", you don't get a polynomial in "y".
Either graph the polynomial on graph paper manually or on a graphing calculator. If it is a "y=" polynomial, then the zeroes are the points or point where the polynomial touches the x-axis. If it is an "x=" polynomial, then the zeroes are the points or point where the polynomial touches the y-axis. If it touches neither, then it has no zeroes.
It is: y^2 -13y +12 = (y-1)(y-12) when factored
A polynomial with a degree of one, of the form y = ax + b, where a and b are constants.
YES! A polynomial is made up of terms that are only added, subtracted or multiplied. Division is not allowed. It may have constants, exponents or variables. A monomial is one type of polynomial. y+2y is the same as 4y which is a monomial. It has degree 1 since the exponent on y is 1.
For a polynomial of the form y = p(x) (i.e., some polynomial function of x), having a y-intercept simply means that the polynomial is defined for x = 0 - and a polynomial is defined for any value of "x". As for the x-intercept: from left to right, a polynomial of even degree may come down, not quite reach zero, and then go back up again. A simple example is y = x2 + 1. Why is the situation for "x" and for "y" different? Well, the original equation is a polynomial in "x"; but if you solve for "x", you don't get a polynomial in "y".
It is a polynomial in x and y.
no...
Either graph the polynomial on graph paper manually or on a graphing calculator. If it is a "y=" polynomial, then the zeroes are the points or point where the polynomial touches the x-axis. If it is an "x=" polynomial, then the zeroes are the points or point where the polynomial touches the y-axis. If it touches neither, then it has no zeroes.
It is: y^2 -13y +12 = (y-1)(y-12) when factored
Yes, 18y3 + 2y2 + 1 is a polynomial; it is a cubic expression. If it were expanded to form an equation, then it would be a cubic equation (or higher), capable of solution.
You set x = 0 and evaluate the polynomial. Note that this should be "y-intercept" in the singular, not in the plural.
The degree for 6xy to the 3rd power is equal to the addition of the exponents of equal polynomial that means 1+3 (1 for the x and 3 for the y) and you get an answer of a 4th degree polynomial
A polynomial with a degree of one, of the form y = ax + b, where a and b are constants.
It is a polynomial of degree one in x, and also a polynomial of degree one in y.
Not quite. The polynomial's linear factors are related - not equal to - the places where the graph meets the x-axis. For example, the polynomial x2 - 5x + 6, in factored form, is (x - 2) (x - 3). In this case, +2 and +3 are "zeroes" of the polynomial, i.e., the graph crosses the x-axis. That is, in an x-y graph, y = 0.