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tigry1 [53]
3 years ago
14

Let f(x)=x^2+2x+3 . What is the average rate of change for the quadratic function from x=−2 to x = 5?

Mathematics
1 answer:
miskamm [114]3 years ago
5 0
To find the average rate of change of  given function f(x) on a given interval (a,b):

Find f(b)-f(a), b-a, and then divide your result for f(b)-f(a) by your result for b-a:


f(b) - f(a)
------------
    b-a

Here your function is f(x) = x^2 - 2x + 3.  Substituting b=5 and a=-2,
f(5) = 5^2 -2(5)+3 =?                          and f(-2) = (-2)^2 - 2(-2) + 3 = ?


Calculate           f(5) - [ f(-2) ]
                         ------------------ using your results, above.
                              5 - [-2]

Your answer to this, if done correctly, is the "average rate of change of the function f(x) = x^2+2x+3 on the interval [-2,5]."
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MAXImum [283]

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Find the value of 3/4 + 1/5 cannnnnn anyone helppp!???
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Value of 3/4 + 1/5 is 19/20.

Do you need the steps?

First, calculate the LCM of the denominators (4 and 5) which is 20.

Now, divide 20 by 4 and 5 and multiply the quotient with the numerator.

=>

<u> </u><u>(</u><u>5</u><u>×</u><u>3</u><u>)</u><u> </u><u>+</u><u> </u><u>(</u><u>4</u><u>×</u><u>1</u><u>)</u><u> </u><u> </u>

20

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20

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8 0
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Y=-1/3x^2-4x-5 in vertex form
grigory [225]

Answer:

Y=-\frac{1}{3}(x+6)^2+7   [Vertex form]

Step-by-step explanation:

Given function:

Y=-\frac{1}{3}x^2-4x-5

We need to find the vertex form which is.,

y=a(x-h)^2+k

where (h,k) represents the co-ordinates of vertex.

We apply completing square method to do so.

We have  

Y=-\frac{1}{3}x^2-4x-5

First of all we make sure that the leading co-efficient is =1.

In order to make the leading co-efficient is =1, we multiply each term with -3.

-3\times Y=-3\times\frac{1}{3}x^2-(-3)\times4x-(-3)\times 5

-3Y=x^2+12x+15

Isolating x^2 and x terms on one side.

Subtracting both sides by 15.

-3Y-15=x^2+12x-15-15

-3Y-15=x^2+12x

In order to make the right side a perfect square trinomial, we will take half of the co-efficient of x term, square it and add it both sides side.  

square of half of the co-efficient of x term = (\frac{1}{2}\times 12)^2=(6)^2=36

Adding 36 to both sides.

-3Y-15+36=x^2+12x+36

-3Y+21=x^2+12x+36

Since x^2+12x+36 is a perfect square of (x+6), so, we can write as:

-3Y+21=(x+6)^2

Subtracting 21 to both sides:

-3Y+21-21=(x+6)^2-21

-3Y=(x+6)^2-21

Dividing both sides by -3.

\frac{-3Y}{-3}=\frac{(x+6)^2}{-3}-\frac{21}{-3}

Y=-\frac{1}{3}(x+6)^2+7   [Vertex form]

8 0
3 years ago
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