Which values of n degree polynomials must there definitely exist an x-intercept. Please explain all the reasoning
2 answers:
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.
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G=-6
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Answer:
x=5
Step-by-step explanation:
x=5 hope this helps!
Answer:
given us,
n=4
here
3^-n+1-3^-n+2
3^-4+1-3^-4+2
-4+1-4+2
-5
Step-by-step explanation:
we can put value in n .we can power add and subtract then we can cut 3 then power we can add or subtract
Answer:
(1,6)
Step-by-step explanation:
all u gotta do is look at the graph, the school is at x=1 and y=6 so therefore (1,6)
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