+2441 I recommend finding the vertex because doing this should make the process easier. This inequality is much harder than the previous one, but do not fret! It is still perfectly manageable by hand. You just need some patience.
As a quick refresher, for a quadratic in the form \(ax^2+bx+c\), the coordinate of the vertex is located at \((\frac{-b}{2a},f(\frac{-b}{2a}))\). Let's figure those coordinates right now!
| \(-2x^2-8x-12\) | This is the original expression. |
| \(a=-2;b=-8\\ x=\frac{-(-8)}{2*-2}=\frac{8}{-4}=-2\) | This work shows the x-coordinate of the vertex. Plug in this x-coordinate to determine the y-coordinate of the vertex. |
| \(x=-2\\ y=-2(-2)^2-8(-2)-12\) | Let's just simplify this. The calculations should not be too difficult to do by hand. |
| \(y=-2(-2)^2-8(-2)-12\\ y=-2*4+16-12\\ y=-8+16-12\\ y=8-12\\ y=4\) | |
| \((-2,-4)\) | This is the coordinate of the vertex. |
The vertex, after all, is a coordinate of the parabola, so we can pencil this coordinate onto the graph. Let's think of x-coordinates that are generally friendly to plug in. x=0 and x=1 seem like excellent candidates. When graphing any parabola, I generally strive to graph at least 5 points. Graphing at least 5 points allows me to understand the general shape of the curve.
Let's calculate the output when x=0:
| \(x=0\\ y=-2x^2-8x-12\) | Replace all instances of x with 0 to find the corresponding y-coordinate. A lot of cancellation will occur here. This will ease the process. If given the choice, always choose x=0 as a point to plug in for parabolas. It is very easy. |
| \(y=-2*0^2-8*0-12\\ y=-12\) | |
| \((0,-12)\) | This is another point of the parabola. Plot this. |
Let's calculate the output when x=1:
| \(x=1\\ y=-2x^2-8x-12\) | Do the same process as before. It is not as easy to substitute in as 1, but it should be relatively easy. |
| \(y=-2*1^2-8*1-12\\ y=-2-8-12\\ y=-22\) | |
| \((1,-22)\) | Graph this coordinate as well. |
Remember the vertex you found earlier? This is where that calculation becomes handy. A vertical line from the vertex has a special name for quadratics: the axis of symmetry. The axis of symmetry indicates that the graph should be symmetrical about this particular line. We know that the vertex is at \((-2,-4)\). I know that \((0,-12)\) is a point, so \((-4,-12)\) is also a point on the parabola. Using the same logic for \((1,-22)\), \((-5,-22)\) is also a point. Without knowing the vertex, this shortcut would be impossible. Now just shade all the points lower than these outputs. You are done.
+2441 MrPatel, this particular problem is not as labor-intensive as you probably think. As the inequalities become more complicated, though, this process will probably not be as simple. I would recommend treating the inequality \(y>2x^2\) as if it were an equation for now. This should help you get started. I will make a table of values with 5 x-values. This should not be too difficult.
| Input (x-coordinate) | Output (y-coordinate) | Coordinate | |
| \(-2\) | \(8\) | \((-2,8)\) | |
| \(-1\) | \(2\) | \((-1,2)\) | |
| \(0\) | \(0\) | \((0,0)\) | |
| \(1\) | \(2\) | \((1,2)\) | |
| \(2\) | \(8\) | \((2,8)\) | |
Plot these coordinates on a coordinate plane. Then, connect them with a dotted curve. Do this because the inequality in a less-than type. The original inequality indicates that we want to include all outputs that are greater than the output given by 2x^2. For example, a valid output for the given equality for the input 0 is 1 or 2 or \(\frac{5}{12}\) or \(\sqrt{30}\) or \(10.123\). Notice how these are all greater than 0. This means that we want to shade all values above the given outputs.
Does this make sense to you?
