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Sagot :
Using quadratic function concepts, it is found that the true statements about the equation are:
- The graph of the quadratic equation has a minimum value.
- The extreme value is at the point (7,-3).
- The solutions are [tex]3 \pm \sqrt{7}[/tex].
What is a quadratic function?
A quadratic function is given according to the following rule:
[tex]y = ax^2 + bx + c[/tex]
- If a > 0, it has a maximum value, and if a < 0, it has a minimum value.
- The extreme value is [tex](x_v,y_v)[/tex], in which:
[tex]x_v = -\frac{b}{2a}[/tex]
[tex]y_v = -\frac{\Delta}{4a}[/tex]
[tex]\Delta = b^2 - 4ac[/tex]
- The solutions are:
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a}[/tex]
[tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a}[/tex]
In this problem, the function is:
[tex]f(x) = x^2 - 6x + 2[/tex]
The coefficients are [tex]a = 1, b = -6, c = 2[/tex].
Since a > 0, the graph has a minimum value.
For the extreme value, we have that:
[tex]x_v = -\frac{b}{2a} = \frac{6}{2} = 3[/tex]
[tex]\Delta = b^2 - 4ac = (-6)^2 - 4(1)(2) = 28[/tex]
[tex]y_v = -\frac{\Delta}{4a} = -\frac{28}{4} = -7[/tex]
Hence:
The extreme value is at the point (7,-3).
The solutions are:
[tex]x_1 = \frac{6 + \sqrt{28}}{2} = \frac{6 + 2\sqrt{7}}{2} = 3 + \sqrt{7}[/tex]
[tex]x_2 = \frac{6 - \sqrt{28}}{2} = \frac{6 - 2\sqrt{7}}{2} = 3 - \sqrt{7}[/tex]
The solutions are [tex]3 \pm \sqrt{7}[/tex].
You can learn more about quadratic function concepts at https://brainly.com/question/24737967
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