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### Course: Algebra (all content)>Unit 9

Sal solves a few quadratic inequalities by moving all terms to one side of the inequality and graphing the resulting expression. Created by Sal Khan.

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• How do you solve this inequality x^2+3x is greater or equal to -2. I've tried doing it and got x is greater or equals to -2 and x is greater or equals to -1 but that is no the right answer. Please help
• `x² + 3x ≥ -2`
`=> x² + 3x + 2 ≥ 0`
`=> x² + 2x + x + 2 ≥ 0`
`=> x(x + 2) + 1(x + 2) ≥ 0`
`=> (x + 1)(x + 2) ≥ 0`
`=> x ∈ [-∞, -2]⋃[-1, ∞]`

Stated simply, `x < -2` or `x > -1`.
• I dont understand how to graph Y >X^2+4
• Also, when an inequality is solved for y, you can interpret "y >" as "above the line" and "y <" as "below the line". Just be careful, as a negative coefficient will reverse this logic. When in doubt, test points!
• What point is the axis of symmetry?
• the axis of symmetry is the line that goes through the vertex. hope this helps! :-)
• Why is the vertex -6 instead of -b/2a or - 1/2 ?
• Sal tells you at about that -6 is the y-intercept (the point (0, -6). He then explicitly tells you that it is not the vertex. The point is close to the vertex, but it is not the vertex.

You can find the x-value of the vertex can be found using -b/2a = -1/2.
Then, you can find the y-value of the vertex by plugging in x = -1/2
• please explain open interval and closed interval and also modulus of real number
• In quadratic inequalities, when do you use dotted/solid lines?
(1 vote)
• Dotted lines: Strict Inequality: ` > or <`
Solid lines: Not Equal/Equal: ` ≥ or ≤`
Hope this helps!
—CT-2/002-24
• How would you actually graph a quadratic inequality? For instance, would you just shade everything outside of or inside of the roots? This part sort of confused me because the graph at was a bit unclear.
• If a parabola is supposed to be symetric, how is the first example valid? Shouldn't the minimum be at the center of the two roots?
(1 vote)
• If you're referring to Sal's graph, it's not drawn particularly to scale. The minimum point is indeed at the midpoint of the two roots `x=-3` and `x=2`. Therefore the vertex should occur at `x=-1/2` (at which point `y=-25/4`). The graph does have a y-intercept of `(0,-6)`, which was the point Sal was using to graph the curve, but it's not the minimum of the function. Good eye! :)