inequality
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!
Suppose your inequality is $P(x) < 0$. Let $r_1 < \ldots < r_k$ be the distinct real roots of $P(x)$, with corresponding multiplicities $d_1, \ldots, d_k$. Thus $P(x) = (x - r_1)^{d_1} (x - r_2)^{d_2} \ldots (x - r_k)^{d_k} Q(x)$ where $Q(x)$ has the same sign for all $x$ (the sign of the leading coefficient of $P$). As $x$ increases or decreases, the sign of $P(x)$ changes at $x = r_j$ if $d_j$ is odd but stays the same if $d_j$ is even. To find the $r_j$, you generally will need to use numerical methods.
In your example, the roots happen to be integers $-2,-3,-4$ and the multiplicities are all $1$. Obviously $P(x) > 0$ for $x > 0$, and thus also for $x > -2$. At each root the sign changes, so $P(x) < 0$ for $-3 < x < -2$, $P(x) > 0$ for $-4 < x < -3$, and $P(x) < 0$ for $x < -4$.
On the other hand, if it had been $R(x) = {x}^{4}+12\,{x}^{3}+53\,{x}^{2}+102\,x+72 = (x + 3) P(x)$, the roots would be the same except that the multiplicity of the root $-3$ is $2$ instead of $1$. Then at $-3$ the sign doesn't change, so $R(x) < 0$ for both $-3 < x < -2$ and $-4 < x < -3$, and $R(x) > 0$ for $x < -4$.
I am giving you a very basic way to find out the regions graphically. For $$y\ge(x-1)^2$$ Note that $y=(x-1)^2$ is a polynomial which has $\mathbb R$ as domain and obviously $\mathbb R_{\ge 0}$ as its range. To find out what that inequality tells you, you can pick two points in and out of the region the parabola made in $\mathbb R^2$. As you see $P$ is in and $Q$ is out with respect the parabola. Now satisfy the coordinates of $P$ into the inequality. You see $$(1/2-1)^2=1/4$$ and it is smaller than $2$. The same for $Q$ tells us $$(3-1)^2=4$$ is greater than $1$. So, the desired region satisfying $$y\ge(x-1)^2$$ is the parabola an the region which is enclosed by it.
Best Answer
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.