Solve the following quadratic inequalities by graphing the corresponding function Looking at these questions and I am not confident in my abilities to solve them.

Solve the following quadratic inequalities by graphing the corresponding function. Note: a) and b) are separate questions.
  $$ a) y \le -2x^2+16x-24\\
b) y > \frac 13 (x-1)^2-3
$$

Help would be appreciated!
 A: For this kind of "simple" inequalities you can use simple concept of higher-lower points. For example, for the first inequality
$$
y \le -2x^2+16x-24
$$
you take all points on the parabola and take any points below that point (since it's less-or-equal) as a solution. If you perform it for all points you'll get the following graph

Same for the second inequality, but now you need to take points above the point on parabola, since it has opposite sign.
$$
y > \frac 13 (x-1)^2-3
$$
Also, it's strict, so points on the parabola are not solutions. I made parabola dashed to emphasize that fact.

A: Step one:
Learn to draw graphs of the equalities.
Suppose we are given
$$
y \le -2x^2+16x-24.
$$
The matching equality is
$$
y = -2x^2+16x-24.
$$
We can factorise this, then graph it.
$$
\begin{align}
y &= -2(x^2-8x+12)\\
&= -2(x-6)(x-2)
\end{align}
$$
This means that the roots of the polynomial are at $x=6$ and $x=2$ respectively. The negative sign of the coefficient of the highest power tells us that the parabola is "upside down", compared to the simplest form of the parabola ($y=x^2$).
Now, we put two points on the cartesian plane at $(6,0)$ and $(2,0)$ respectively. For a parabola, we know that the maxima (or minima) will lie exactly half-way between the roots. i.e. when $x=4$. We can find the $y$ value of this maxima by substituting $x=4$ into the equation:
$$
\begin{align}
y &= -2x^2+16x-24\\
&= -2(4)^2+16(4)-24\\
&= 8
\end{align}
$$ 
So the turning point (a maxima in this case) lies at $(4,8)$.
We now have three points of the parabola, $(2,0)$, $(4,8)$, and $(6,0)$, which we can plot as follows:

Now we can join the dots to make a parabola.

Step two:
Solve inequalities using graphs.
To solve the original inequality, we simply check the two regions on either side of the parabola to see whether they make the inequality true or false.
We can choose any point in the region, provided it is not actually on the parabola. Suppose we check the point $(0,0)$, which lies above and to the left of the parabola. Substituting $x=0$, $y=0$ into the inequality, we get
$$
\begin{align}
y &\le -2x^2+16x-24\\
0 &\le -2(0)^2 +16(0) - 24\\
0 &\le -24\\
\end{align}
$$
Of course, this is FALSE, which means that the region above the parabola does not satisfy the inequality.
Testing the region below the inequality, we substitute any point in that region. For instance, $(4,1)$.
$$
\begin{align}
y &\le -2x^2+16x-24\\
1 &\le -2(4)^2 + 16(4) -24\\
1 &\le 8\\
\end{align}
$$
This is evidently TRUE.
This means that the inequality is true for any point in that region below the parabola. (i.e. The shaded region.) 
(I'm stealing @Kaster's image, because Wolfram|Alpha didn't want to play nicely)

We can describe this region using set theory as follows:
$$
\begin{align}
R = \{(x,y) \in \mathbb{R}^2:-\infty \le x \le \infty,\,\, y \le -2x^2 +16x -24\}
\end{align}
$$
However, the simplest way to describe the region without using the graph is to use the inequality given in the question:
$$
y \le -2x^2+16x-24.
$$
Because this inequality defines a region, we can't write it any more concisely than that.
A: You don't solve these inequalities for points on a curve; these statements describe regions of the plane (so you would "color in" a portion of the graph accordingly).
For the first one, we want all the points which are below or on the parabola described by the quadratic polynomial. Since it is a "downward-facing" parabola (the quadratic coefficient is negative), you would select "everything inside" the parabola and the parabola itself.
In the second one, you would only have the points above, but not on, the parabola (since this is a "proper inequality" -- no equation here).  The parabola "faces upward" in this problem (the quadratic coefficient is positive), so again you would "color in" all the points within the parabola, but not the parabola itself.
