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The table of values represents a quadratic function.


What is the the average rate of change for f(x) from x=−5 to x = 10 ?



Enter your answer in the box.


x ​f(x)​

​−10​ 184

​−5​ 39

0 ​−6​

5 49

10 204

Respuesta :

[tex]\bf \begin{array}{rrllll} x&f(x)\\ \textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\ -10&184\\ \boxed{-5}&\boxed{39}\\ 0&-6\\ 5&49\\ \boxed{10}&\boxed{204} \end{array} \\\\\\ \\\\\\ \cfrac{f(x_2)-f(x_1)}{x_2-x_1}\impliedby \textit{average rate of change} \\\\\\ \cfrac{f(10)-f(-5)}{10-(-5)}\implies \cfrac{f(10)-f(-5)}{10+5}\implies \cfrac{\boxed{204}-\boxed{39}}{10+5} \\\\\\ \cfrac{165}{15}\implies 11[/tex]

Answer:

The average rate of change for f(x) from x=−5 to x = 10 is, 11

Step-by-step explanation:

Average rate A(x) of change for a function f(x) over [a, b] is given by:

[tex]A(x) = \frac{f(b)-f(a)}{b-a}[/tex]

As per the statement:

We have to find the average rate of change for f(x) from x=−5 to x = 10.

From the table we have;

At x = -5

f(-5) = 39

and

at x = 10

f(10) = 204

Substitute these in [1] we have;

[tex]A(x) = \frac{f(10)-f(-5)}{10-(-5)}[/tex]

⇒[tex]A(x) = \frac{204-39}{10+5}[/tex]

⇒[tex]A(x) = \frac{165}{15}[/tex]

Simplify:

A(x) = 15

Therefore, the average rate of change for f(x) from x=−5 to x = 10 is, 11