1) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is all real numbers because it asked you to solve y>0 for x.
2) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is all real numbers because it asked you to solve y>=0 for x.
3) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is none because we are looking for when y<0 for x. (y<0 means y is negative)
4) Parabola is opened up with no x-intercepts so the graph is fully above the x-axis which means all the y-coordinates in our points are positive so that is why the solution is none because we are looking for when y<=0 for x. (y<=0 means negative or 0)
Step-by-step explanation:
![y=ax^2+bx+c[/tex forms a parabola when graph assuming [tex]a \neq](/tpl/images/0083/9256/b91fa.png)
.
If 
, we would have a parabola.
Anyways here are few things that might help when solving these:
1) Write ax^2+bx+c in factored form; it can lead to the x-intercepts quickly once you have it
2) The discriminant b^2-4ac:
A) It tells you if you have two x-intercepts if b^2-4ac is positive
B) It tells you if you have one x-intercept if b^2-4ac is zero
C) It tells you if you have zero x-intercepts if b^2-4ac negative
3) If a>0, then the parabola opens up.
If a<0, then the parabola opens down.
4) You might choose to test before and after the x-intercepts too, using numbers between or before and after.
Let's look at questions labeled (1)-(4).
(1) x^2-x+2>0
a=1
b=-1
c=2
I'm going to use the discriminant to see how many x-intercepts I have:
b^2-4ac
(-1)^2-4(1)(2)
1-8
-7
Since -7 is negative, then we have no x-intercepts.
The parabola is also opened up.
So if we have no-x-intercepts and the parabola is opened up (a is positive), then the parabola is above the x-axis.
So the y values for x^2-x+2 where y=x^2-x+2 is positive for all real numbers.
The solution to x^2-x+2>0 is therefore all real numbers.
(2) x^2+5x+7>=0
a=1
b=5
c=7
b^2-4ac
(5)^2-4(1)(7)
25-28
-3
Since -3 is negative, we have no x-intercepts.
The parabola is also opened up because a=1 is positive.
So again x^2+5x+7>=0 has solutions all real numbers.
(3) x^2-4x+5<0
a=1
b=-4
c=5
b^2-4ac
(-4)^2-4(1)(5)
16-20
-4
Since -4 is negative, we have no x-intercepts.
The parabola is opened up because a=1 is positive.
All the y-coordinates for our points are positive.
So y=x^2-4x+5<0 has no solutions because there are no y's less than 0.
(4) x^2+6x+10<=0
a=1
b=6
c=10
b^2-4ac
6^2-4(1)(10)
36-40
-4
Since -4 is negative, the parabola has no x-intercepts.
The parabola is opened up because a=1 is positive.
All the y-coordinates on our parabola are positive.
So y=x^2+6x+10<=0 has no solutions because there are no y's less than 0 or equal to 0.
