The value of [tex]x[/tex] associated to the extreme value of the quadratic equation [tex]k(x) = (300+10\cdot x)\cdot (5-0.2\cdot x)[/tex]. The extreme value is [tex]k\left(-\frac{5}{2} \right) = \frac{3025}{2}[/tex].
First, we need to expand the polynomial as follows:
[tex]k(x) = (300+10\cdot x)\cdot (5-0.2\cdot x)[/tex]
[tex]k(x) = 300\cdot (5-0.2\cdot x)+10\cdot x\cdot (5-0.2\cdot x)[/tex]
[tex]k(x) = 1500-60\cdot x +50\cdot x -2\cdot x^{2}[/tex]
[tex]k(x) = -2\cdot x^{2}-10\cdot x +1500[/tex]
[tex]-2\cdot x^{2}-10\cdot x +[1500-k(x)] = 0[/tex]
By the quadratic formula, we know that a value of [tex]x[/tex] pass through the axis of symmetry of the parabola if and only if the discriminant is zero:
[tex]-100-4\cdot (-2)\cdot [1500-k(x)] = 0[/tex]
And the value of [tex]x[/tex] is:
[tex]x = -\frac{(-10)}{2\cdot (-2)}[/tex]
[tex]x = -\frac{5}{2}[/tex]
And the extreme value is:
[tex]k\left(-\frac{5}{2} \right) = -2\cdot \left(-\frac{5}{2} \right)^{2}-10\cdot \left(-\frac{5}{2} \right) + 1500[/tex]
[tex]k\left(-\frac{5}{2} \right) = \frac{3025}{2}[/tex]
The value of [tex]x[/tex] associated to the extreme value of the quadratic equation [tex]k(x) = (300+10\cdot x)\cdot (5-0.2\cdot x)[/tex]. The extreme value is [tex]k\left(-\frac{5}{2} \right) = \frac{3025}{2}[/tex].
We kindly invite to check this question on symmetry: https://brainly.com/question/17759737
