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The minima of the quadratic function is at the vertex, it is at (2, -4).
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).
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The exact location of all the relative and absolute extrema of the function is (2, -4).
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