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vitfil [10]
2 years ago
11

A bakery is collecting data to investigate how changing the price charged for a loaf of bread affects the bakery’s daily profit.

The graph shows the data the bakery has collected. Which quadratic function is the best model for the data? y = − 8 ( x + 4 ) 2 + 800 y = − 8 ( x + 4 ) 2 + 800 y = − 8 ( x − 4 ) 2 + 800 y = − 8 ( x − 4 ) 2 + 800 y = − 200 ( x + 4 ) 2 + 800 y = − 200 ( x + 4 ) 2 + 800 y = − 200 ( x − 4 ) 2 + 800
Mathematics
1 answer:
RSB [31]2 years ago
6 0

Considering the vertex of the graph, it is found that the quadratic function that is the best model for the data is given by:

y = -200(x - 4)^2 + 800

<h3>What is the equation of a quadratic function given it’s vertex?</h3>

The equation of a quadratic function, of vertex (h,k), is given by:

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

In which a is the leading coefficient.

Researching the problem on the internet, it is found that the vertex is at point (4,800), hence h = 4 and k = 800.

y = a(x - 4)^2 + 800

It has a value of y = 0 at x = 6, hence:

0 = a(6 - 4)^2 + 800

a = -200

Thus, the model is:

y = -200(x - 4)^2 + 800

More can be learned about quadratic functions at brainly.com/question/24737967

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Convert 385 pounds into kg given the fact that 1 kg = 2.2 pounds.

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Part 1) Intervals of the domain where the  graph is above the x-axis (f(x) > 0)

Part 2) location on graph where input is zero  (y-intercept)

Part 3) location on graph where output is zero  (x-intercept)

Part 4) Intervals of the domain where the  graph is below the x-axis (f(x) < 0)

Step-by-step explanation:

<u><em>Verify each case</em></u>

Part 1) we have

Intervals of the domain where the  graph is above the x-axis

we know that

If the graph is above the x-axis, then the value of f(x) is positive

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f(x) > 0

Part 2) we have

location on graph where input is zero  

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x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The y-intercept is the value of f(x) (output value) when the value of x (input value) is zero

therefore

y-intercept

Part 3) we have

location on graph where output is zero  

Let

x ---> the independent variable or input value

f(x) ---> the dependent variable or output value

we know that

The x-intercept is the value of x (input value) when the value of the function f(x) (output value) is zero

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Part 4) we have

Intervals of the domain where the  graph is below the x-axis

we know that

If the graph is below the x-axis, then the value of f(x) is negative

therefore

f(x) < 0

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3 years ago
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