You set x = 0 and evaluate the polynomial. Note that this should be "y-intercept" in the singular, not in the plural.
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For a line, this is the x-intercept. For a polynomial, these points are the roots or solutions of the polynomial at which y=0.
No. An expression can have a variable exponent (for instance, 2 to the power x, or x to the power y), but that is no longer a polynomial.
if you make a t chart with the x and y, you can find functions. the change in y is the number in front of the x (multiplying) and if you find the x at 0, the y will be what you add. here: y=3x + 1 the 3 is the change in y the + 1 when you have (0, 1) as a point. subistite the axis of symmetry into the original problem and solve for y. inentify the vertex
The question cannot be answered without information about the nature of the curve, for example, what degree polynomial (if it is a polynomial).
You can factor this multivariate polynomial (a polynomial with several variables, here x and y), by looking at it as a univariate polynomial in either x or y. This would give you a simple second order equation of the form ax2+bx+c, which you can solve. This will give you 2 solutions, say x1 and x2, and can then factor your polynomial to a(x-x1)(x-x2). In our case: a=8, b=-y and c=-7y2 and the solutions for this equation are x1=y and x2=-7/8*y and this gives us 8x2 - xy -7y2 = 8(x-y)(x+7/8y) = (x-y)(8x+7y)