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-BARSIC- [3]
2 years ago
15

What are the possible degrees for the polynomial function

Mathematics
2 answers:
Anit [1.1K]2 years ago
7 0

Answer:

Degree of the polynomial function = 5

Step-by-step explanation:

Since, the given graph of a polynomial has the x-intercepts at x = -2, -0.5, 0.5, 1.5, 3

Therefore, the equation of the polynomials with zero roots will be,

y = (x + 2)(x + 0.5)(x - 0.5)(x - 1.5)(x - 3)

When we solve this expression highest degree of the variable x will be 5.

Therefore, the possible degree of the polynomial function = 5

scoray [572]2 years ago
5 0

Answer:

i was going to screenshot the answer bc i got it right but its

"odd 5 degrees or greater" :)

hope all you guys are healthy and safe

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Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s
Nataliya [291]

Answer:

f(x)=\frac{1}{2}x^2-4x+5

Step-by-step explanation:

A parabola is written in the form

f(x)=a((x-h)^2+k) (1)

where:

h is the x-coordinate of the vertex of the parabola

ak is the y-coordinate of the vertex of the parabola

a is a scale factor

For the parabola in the problem, we know that the vertex has  coordinates (4,-3), so we have:

h=4 (2)

ak=-3

From this last equation, we get that a=\frac{-3}{k} (3)

Substituting (2) and (3) into (1) we get the new expression:

f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3 (4)

We also know that the parabola  contains the point (2,-1), so we can substitute

x = 2

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Into eq.(4) and find the value of k:

-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6

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a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}

So the equation of the parabola is:

f(x)=\frac{1}{2}((x-4)^2 -6) (5)

Now we want to rewrite it in the standard form, i.e. in the form

f(x)=ax^2+bx+c

To do that, we simply rewrite (5) expliciting the various terms, we find:

f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5

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