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zysi [14]
2 years ago
5

Find the range of the quadratic function. (y=2x^2-20x+53)

Mathematics
1 answer:
Anit [1.1K]2 years ago
4 0

Answer: The range is y ≥ 3.

Step-by-step explanation:

When we have an equation:

y = f(x)

The range is the set of the possible values of y.

In this case, we have:

y = f(x) = 2*x^2 - 20*x + 53

This is a quadratic equation, and the leading coefficient is positive. This means that the arms of the graph will open upwards.

Then the minimum value of y will be the vale at the vertex.

We know that for a quadratic equation of the form:

a*x^2 + b*x + c

The vertex is at:

x = -b/2a

Then in this case, the vertex will be at:

x = -(-20)/(2*2) = 20/4 = 5

Then the smallest value of this function will be:

f(5) = 2*5^2 - 20*5 + 53 = 3

This is the smallest value that y can take, and y can take any value greater than 3.

Then the range can be written as:

y ≥ 3.

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The attached graph represents the graph of f(x) = (x - 1)^2 - 2

<h3>How to plot the graph?</h3>

The equation is given as:

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Next, we set x to -2, -1, 0, 1 and 2.

So, we have:

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This means that the table of values is

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Next, we plot the above points and connect them.

See attachment for the graph of f(x) = (x - 1)^2 - 2

Read more about graphs and functions at:

brainly.com/question/4025726

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