Hi, I haven't been able to find the answer to this anywhere, so Ill ask it here. I am doing an assignment on quadratic inequalities, and some of the quadratic inequalities are shaded (I know how the shading aspect works), but others are not. As in, they do not have either part of them darkened, they only have the two dots on the x axis that are either hollow or or solid. My question is, why are some not shaded inside or out? Does this have to do with what form the inequality is in (i.e. factored). Any help is appreciated. Thanks, Ben
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Help with quadratic inequalities
- Thread starter Maz
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Hi, I haven't been able to find the answer to this anywhere, so Ill ask it here. I am doing an assignment on quadratic inequalities, and some of the quadratic inequalities are shaded (I know how the shading aspect works), but others are not. As in, they do not have either part of them darkened, they only have the two dots on the x axis that are either hollow or or solid. My question is, why are some not shaded inside or out? Does this have to do with what form the inequality is in (i.e. factored). Any help is appreciated. Thanks, Ben
Please post an example of a problem where you are seeing this. It's hard (at least for me) to understand what you are talking about unless I see the problem. Thanks!
If the form of the inequality was "not equal", \(\displaystyle \ne 0\), then the statement would be true everywhere EXCEPT at the two (or one or 0) points where the curve crosses the x-axis.Hi, I haven't been able to find the answer to this anywhere, so Ill ask it here. I am doing an assignment on quadratic inequalities, and some of the quadratic inequalities are shaded (I know how the shading aspect works), but others are not. As in, they do not have either part of them darkened, they only have the two dots on the x axis that are either hollow or or solid. My question is, why are some not shaded inside or out? Does this have to do with what form the inequality is in (i.e. factored). Any help is appreciated. Thanks, Ben
Please post an example of a problem where you are seeing this. It's hard (at least for me) to understand what you are talking about unless I see the problem. Thanks!
I am given a matching question with 3 inequalities and 3 graphs of parabolas. The inequalities are: A) 2.19(x+2)^2-2=<0 B)2.19(x+2)^2-2>0 C) 2.19x^2+8.76x+6.76.
The graphs I am given are all the same parabola, but one of them is shaded on the inside of the parabola, and the other two are not shaded at all. Out of those two parabolas, one of them has hollow circles on the x axis, and the other has solid.
Using logic, I reasoned that the shaded one must match C), as C) is the most different from the other two. Then I reasoned that the parabola with solid points must be A), as I know that solid points mean that the inequality is greater/less than or equal to. After doing that, I knew that the graph of B) was the one with the hollow points.
My question is, why are two of the parabolas not shaded inside or out? I am assuming there is a rule that I am unaware of. Thanks again.
Hi, I haven't been able to find the answer to this anywhere, so Ill ask it here. I am doing an assignment on quadratic inequalities, and some of the quadratic inequalities are shaded (I know how the shading aspect works), but others are not. As in, they do not have either part of them darkened, they only have the two dots on the x axis that are either hollow or or solid. My question is, why are some not shaded inside or out? Does this have to do with what form the inequality is in (i.e. factored). Any help is appreciated. Thanks, Ben
As Sir Michael says, it is hard to know what you are asking without an example.
However \(\displaystyle x^2 - 5x + 6 \ne -3x^2 + 15x - 18\) at every value of x except x = 2 and x = 3.
Proof: \(\displaystyle x^2 - 5x + 6 = -3x^2 + 15x - 18 \implies 3x^2 + x^2 - 15x - 5x + 6 + 18 = 0 \implies 4x^2 - 20x + 24 = 0 \implies\)
\(\displaystyle x = \dfrac{20 \pm \sqrt{(-20)^2 - 4(4)(24)}}{2 * 4} =\dfrac{20 \pm \sqrt{400 - 384}}{8} = \dfrac{20 \pm \sqrt{16}}{8} = \dfrac{20 \pm 4}{8} \implies\)
\(\displaystyle x = 2\ or\ x = 3.\)
So dots at (2, 0) and at (3, 0) would represent where the inequality was false.
\(\displaystyle f(x) = 2.19(x+2)^2 - 2 = 2.19(x^2 + 4x + 4) - 2 = 2.19x^2 + 8.76x + 6.76\)I am given a matching question with 3 inequalities and 3 graphs of parabolas. The inequalities are: A) 2.19(x+2)^2-2=<0 B)2.19(x+2)^2-2>0 C) 2.19x^2+8.76x+6.76.
The graphs I am given are all the same parabola, but one of them is shaded on the inside of the parabola, and the other two are not shaded at all. Out of those two parabolas, one of them has hollow circles on the x axis, and the other has solid.
Using logic, I reasoned that the shaded one must match C), as C) is the most different from the other two. Then I reasoned that the parabola with solid points must be A), as I know that solid points mean that the inequality is greater/less than or equal to. After doing that, I knew that the graph of B) was the one with the hollow points.
My question is, why are two of the parabolas not shaded inside or out? I am assuming there is a rule that I am unaware of. Thanks again.
The three parabolas are identical. The two roots are at \(\displaystyle x = -2 \pm \sqrt{2/2.19}\)
A) The domain of \(\displaystyle f(x) \le 0 \) is the portion of the x-axis between the two roots and including the two roots (solid points).
B) The domain of \(\displaystyle f(x) > 0 \) is the portions beyond the two roots, and not including the roots (hollow points).
C) doesn't have an inequality in it, as typed.
\(\displaystyle f(x) = 2.19(x+2)^2 - 2 = 2.19(x^2 + 4x + 4) - 2 = 2.19x^2 + 8.76x + 6.76\)
The three parabolas are identical. The two roots are at \(\displaystyle x = -2 \pm \sqrt{2/2.19}\)
A) The domain of \(\displaystyle f(x) \le 0 \) is the portion of the x-axis between the two roots and including the two roots (solid points).
B) The domain of \(\displaystyle f(x) > 0 \) is the portions beyond the two roots, and not including the roots (hollow points).
C) doesn't have an inequality in it, as typed.
My apologies. I meant to type it as y>=2.19x^2+8.76x+6.76
ok - I would transpose and write that asMy apologies. I meant to type it as y>=2.19x^2+8.76x+6.76
C) \(\displaystyle f(x) \le y\),.... \(\displaystyle f(x)-y \le 0\)
Find the roots of \(\displaystyle f(x)-y\) and treat the result like part (A)