Question

Complete the square to rewrite y-x^2-6x+14 in vertex form. then state whether the vertex is a maximum or minimum and give its cordinates

Answer 1

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:

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

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

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