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Sagot :
Given:
[tex]y=2x^2-8x-10[/tex]
To find:
The interval(s) on which the given function is positive.
Solution:
We have,
[tex]y=2x^2-8x-10[/tex]
[tex]y=2(x^2-4x-5)[/tex]
[tex]y=2(x^2-5x+x-5)[/tex]
[tex]y=2(x(x-5)+1(x-5))[/tex]
[tex]y=2(x+1)(x-5)[/tex]
For zeroes, [tex]y=0[/tex]
[tex]2(x+1)(x-5)=0[/tex]
[tex]x=-1,x=5[/tex]
-1 and 5 divide the number line in three parts [tex](-\infty , -1),(-1,5),(5,\infty )[/tex].
Interval Selected Value [tex]y=2(x+1)(x-5)[/tex] Sign
[tex](-\infty,-1)[/tex] -2 [tex]y=2(-2+1)(-2-5)=14[/tex] +
[tex](-1,5)[/tex] 0 [tex]y=2(0+1)(0-5)=-10[/tex] -
[tex](5,\infty)[/tex] 6 [tex]y=2(6+1)(6-5)=14[/tex] +
So, the function is positive for interval [tex](-\infty,-1)[/tex] and [tex](5,\infty)[/tex].
We can say that the function is positive for interval x < -1 and x > 5.
Therefore, the correct option is B.
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