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viktelen [127]
3 years ago
10

Determine the intervals on which the polynomial is entirely negative and those on which it is entirely positive. (Enter your ans

wers using interval notation. Enter EMPTY or ∅ for the empty set.)
−x^2 + 6x − 10
I don't understand how to factor this, as as such, I can't figure out how to answer the question. Please help!
Mathematics
2 answers:
Stells [14]3 years ago
5 0
This is a parabola that opens downward.
Remember that the axis of symmetry is located at x=-b/(2a).
So the vertex is at (3,-1)
So functions slope is positive or increasing on (-co, 3] and decreasing on [3,co).

Law Incorporation [45]3 years ago
3 0

Answer with explanation:

The function for which we have to find , the intervals on which the polynomial is entirely negative and those on which it is entirely positive.

 f(x)= -x²+6 x -10

 If you will find the root of the function, there is no real root.

To find the root we will use Discriminant formula

For a Quadratic function, ax²+b x+c=0,

 x=\frac{-b+\sqrt{D}}{2a}\\\\D=b^2-4ac\\\\x=\frac{-6\pm \sqrt{6^2-4 *(-1)*(-10)}}{2*(-1)}\\\\x=\frac{-6\pm\sqrt{-4}}{-2}

→So,there is no interval in which  the polynomial is entirely negative and those on which it is entirely positive, which can be represented in interval notation using Ф.

   f(x)= -(x^2-6x+10)\\\\= - [(x-3)^2-9+10]\\\\=-(x-3)^2-1

For, any value of x,the value of f(x) will be always Negative.

To find the vertex, put ,x-3=0

 x=3

And, by putting , x= 3 ,in the equation we get

y = -1

So,Vertex = (3, -1)

⇒If you will try to find the intervals in which the function is increasing , means the curve is moving up is from (-∞, 3) and the intervals in which the function is decreasing , means the curve is moving downward is from , (3, ∞).

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Subtract 5/12 × (equation 1) from equation 2:
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please note that the parentheses "{" should span over both equations but the editor doesn't allow that so I should it on both line, See attached example.










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Vika [28.1K]

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