Answer:
There will definitely exist an x-intercept for all n degree polynomials (a·xⁿ + c) that have real roots where n>0 or c = -a
Step-by-step explanation:
The x=intercept of a polynomial is the coordinates of the point on the polynomial where the equation of the polynomial f(x) = 0
Therefore, the coordinates of the x-intercept = (k, 0)
Where x = k is the solution of the f(x) = 0
Hence, there will definitely exist an x-intercept for all n degree polynomials (a·xⁿ + c) that have real roots where n>0 or c = -a.
Answer: There must be definitely at most n value of x - intercept
Step-by-step explanation: If a polynomial has a value of n degree, then, there must be definitely at most n value of x - intercept.
For instance, The polynomial has a degree of 8, so there are at most 8 x-intercepts and at most 8–1 = 7 turning points.
PO
Sin(?)=35/48
?=Sin^-1(35/48)
?=46.82
<u>? ≈ 47</u>
<u>OAmalOHopeO</u>
put them in least to greatest, so that would be: 0,0,1,1,1,1,2,3,3,6,8,7,8, then do the greatest minus the lowest: 8-0=8
