Given $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest? The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest?
$(A)$ $P(-1)$ 
$(B)$ The product of the zeros of $P$
$(C)$ The product of the non-real zeros of $P$ 
$(D)$ The sum of the coefficients of $P$
$(E)$ The sum of the real zeros of $P$

I know that 
$P(-1) = 1-a+b-c+d $ also the product of zeroes is $d$. From the graph the real zeroes are around $1.7$ and $3.85$, so product of non-reals is $\frac{d}{1.7*3.85}$. The sum of the coefficients is $1+a+b+c+d$ and the sum of the zeros is $-a$. Now with this information I don't see immediately how to tell which is the smallest. 
 A: We have, since the non-real root are conjugates  $$P(x)=(x-(m+ni))(x-(m-ni))(x-a)(x-b)$$ or
$$P(x)=x^4-(a+b-2m)x^3+(ab+2m(a+b)+m^2+n^2)x^2-(2abm+(a+b)(m^2+n^2))x+ab(m^2+n^2)$$ It is not hard to compare $m^2+n^2$ which corresponds to $(C)$ above with the other quantities $(A),(B),(D),(E)$.
For example,  calculate a quartic polynomial having a graphic resemblance (this is not immediate!). I found
$$P(x)=x^4-5x^3+\dfrac{153}{64}x^2+\dfrac{181}{32}x+\dfrac{25}{8}$$ so I get the "analogues"$$\begin{cases}(A)=5.859375\\(B)=3.125\\(C)=0.390625\\(D)=7.171875\\(E)=6\end{cases}$$ from which the smallest is $(C)=\dfrac{25}{64}$, i.e. the product of non-real zeros $\dfrac{4\pm3i}{8}$of $P$.
A: Essentially everything can be read from the graph.
(A) is obviously the value at $x=-1$, which is between $4$ and $4.5$.
(B), the product of the roots, is equal to $d$, i.e. $P(0)$, which is $\approx 5.5$.
(C) as you say, is $d$ divided by the two visible zeros (since there is a minimum above the axis in this positive quartic, there are only two), which is $\approx 5.5/(1.7 \times 3.9) < 0.9 $.
(D), the sum of the coefficients, is $1+a+b+c+d=P(1)$, which from the graph is $\approx 3.5$.
(E) is $\approx 1.7+3.9 = 5.6$
Whence (C) is definitely smallest.
