A 2-column table with 5 rows. The first column is labeled x with entries negative 2, negative 1, 0, 1, 2. The second column is labeled f of x with entries 3, negative 2, negative 3, 0, 7.
What is the rate of change for the interval between 0 and 2 for the quadratic equation as f(x) = 2x2 + x – 3 represented in the table?

one-fifth
4
5
10

Answers

Answer 1

The rate of change for the interval between 0 and 2 for the quadratic equation f(x) = 2x^2 + x - 3, represented in the table, is 6.

To find the rate of change for the interval between 0 and 2 for the quadratic equation f(x) = 2x^2 + x - 3, we can use the values provided in the table.

First, let's calculate the values of f(x) for x = 0 and x = 2:

For x = 0: f(0) = 2(0)^2 + 0 - 3 = -3

For x = 2: f(2) = 2(2)^2 + 2 - 3 = 9

Now, we can find the rate of change by calculating the difference in the values of f(x) divided by the difference in x for the interval [0, 2]:

Rate of change = (f(2) - f(0)) / (2 - 0)

Substituting the values we found:

Rate of change = (9 - (-3)) / (2 - 0)

= (9 + 3) / 2

= 12 / 2

= 6

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Answer:

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SOLUTION -;

[tex] 1)\displaystyle{} \: \sqrt{4} = \sqrt{2 \times 2} = \sqrt{2} \\ \: \: \: \bold{irrational \: number}[/tex]

[tex]2) \displaystyle{} \: \sqrt{3} = \bold{irrational \: number}[/tex]

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[tex] 3) \: \: \displaystyle \frac{3}{5} \bold{irrational \: number}[/tex]

[tex] \displaystyle \: \frac{3}{5} \: \bold{rational \: number}[/tex]

[tex] \displaystyle{}\frac{1}{4} \bold{rational \: number}[/tex]

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