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weeeeeb [17]
3 years ago
15

Determined if the graph represents a polynomial function if it does determine the number of turning points and the least degree

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
7 0
  • <em>The graph of f is a polynomial function</em>
  • <em>There are two turning points, namely, x = 0  and x = -2</em>

<h2>Explanation:</h2>

A polynomial function in one variable is given by the form:

f(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots +a_{2}x^2+a_{1}x+a_{0}

Since you haven't provided any expression, I'll choose the following function:

f(x)=x^{3}+3x^{2}+3

So this is indeed a polynomial function. A turning point is an x-value where we have either a local maximum or local minimum. So we need to take the derivative of this functions:

f'(x)=3x^2+6x \\ \\ Finding \ turning \ points: \\ \\ 3x^2+6x=0 \\ \\ x(3x+6)=0 \\ \\ \\ So: \\ \\ x=0 \\ \\  and \\ \\ \ 3x+6=0 \therefore \ \boxed{x=-2}

Conclusion:

  • <em>The graph of f is a polynomial function</em>
  • <em>There are two turning points, namely, x = 0  and x = -2</em>

<em />

<h2>Learn more:</h2>

Polynomial function: brainly.com/question/13729121

#LearnWithBrainly

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Step-by-step explanation:

If a quadratic function has roots 1 and 5

f(x) = (x -1)(x- 5)

f(x) = x^2 - 6x + 5

Unless you meant.  -4 and 6  ?

g(x) = (x + 4)(x - 6)

g(x) = x^2 -2x -24

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Or did you mean  x = 1 and x =4 ?...

x^2 + 2x + 4 = 0  :   complete square  x^2 + 2x + 1 + 3 = 0,   (x+1)^2 + 3 = 0

x^2 - 2x + 4 = 0 :  complete square:  (x -1)^2 + 3 = 0

0x^2 + 2x - 4 = 0,    2x - 4 = 0,  x = 2

x^2 - 2x - 4 = 0  becomes:   x^2 - 2x + 1 - 1 -4 = 0 ;  (x - 1)^2 - 5 = 0

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4 years ago
Read 2 more answers
The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is
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Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.

An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.



Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.

Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.

Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.



Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.

To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.

For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false

For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true

For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true

For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false

For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false

For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.
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