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Find the exact location of all the relative and absolute extrema of the function. (order your answers from smallest to largest x. ) g(x) = x2 − 4 x

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The minima of the quadratic function is at the vertex, it is at  (2, -4).

Where are the relative minima/maxima of the function?

First, we can see that our function is a quadratic function of positive leading coefficient. So it opens upwards.

Then we will only have a minima, which is located in the vertex of the parabola.

Remember that for:

y = a*x^2 + b*x + c

The x-value of the vertex is h = -b/2a

Then for g(x) = x^2 - 4x we have:

h = -(-4)/2*1 = 2

And the y-value of the vertex is:

g(2) = 2^2 - 4*2 = -4

Meaning that the minima of the function g(x) is (2, -4).

If you want to learn more about quadratic functions:

https://brainly.com/question/1214333

#SPJ1

The exact location of all the relative and absolute extrema of the function is (2, -4).

Where are the relative minima/maxima of the function?

First, we can see that our function is a quadratic function of the positive leading coefficient.

Then we will only have a minimum, which is located in the vertex of the parabola.

The equation of the parabola is;

[tex]\rm y = ax^2 + bx + c[/tex]

The x-value of the vertex is h = -b/2a

Then for g(x) = x^2 - 4x we have:

h = -(-4)/2*1 = 2

And the y-value of the vertex is:

g(2) = 2^2 - 4*2 = -4

Meaning that the minima of the function g(x) are (2, -4).

Hence, the exact location of all the relative and absolute extrema of the function is (2, -4).

Learn more about quadratic functions:

brainly.com/question/1214333

#SPJ4

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