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The intercepts of y = − x^2+ b x + c y = - x 2 + b x + c are − 2 ± √ 17 - 2 ± 17 . Which of the following are solutions of − x^2 + b x + c ≥ 0 - x 2 + b x + c ≥ 0 ?

The intercepts of y x2 b x c y x 2 b x c are 2 17 2 17 Which of the following are solutions of x2 b x c 0 x 2 b x c 0 class=

Respuesta :

Given a generic parabola

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

the parabola is concave up if [tex] a>0 [/tex] and concave down if [tex] a<0 [/tex]

Since in this case [tex] a=-1[/tex], the parabola is concave down, which means that it is positive between its solutions.

Since

[tex] -2-\sqrt{17} \approx -6.12,\quad -2+\sqrt{17} \approx 2.12 [/tex]

all options between -6.12 and 2.12 will be fine.

So, the first three options fall inside this interval, and thus f(x)>0 for those points.

Instead, [tex] \sqrt{17} [/tex] is greater than the greater solution, and so f(x)<0 for that point.

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