2 years ago

I need help with this

Mathematics

1 answer:

Bas_tet [7]2 years ago

3 0

It is the square of the input value

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Answer:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

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The vertex for this case would be:

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And the minimum for the function would be 3 and there is no maximum value for the function

Step-by-step explanation:

For this case we have the following equation given:

$y= x^2 -6x +14$

We can complete the square like this:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

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