Derivatives Problem: Why is second term's coefficient less than zero? Problem:

If $f(x)=cx^2+dx+e$ for the function shown in the graph, then what values can $c$, $d$, and $e$ take on? 

The answer is $c<0$, $d<0$, $e>0$. I just don't understand why $d<0$. Since the slope of $f(x)$ is always negative, $f'(x)<0$ and since $f'(x)=(2c)x+d$, $(2c)x+d<0$. Since $c<0$, let's choose $c=-1$, and let's choose $x=3$. $(2)(-1)(3)+d<0$ which leads to $-6+d<0$ which leads to $d<6$. And the maximum value (in this case 6) of this open inequality can change based on what $c$ and $x$ values are chosen. So where am I wrong?
 A: The $x$-coordinate ( $\frac{-d}{2c}$ ) of the parabola vertex must be negative (to the left of the y-axis).
We know $c$ must be negative, because the parabola opens downwards.
So $d$ must be negative. 
A: $$f'(x)=2cx+d \implies f'(0)=d.$$
The slope at $x=0$ is negative, therefore $d<0$.
A: No, for a quadratic polynomial, the slope  doesn't have a constant sign (otherwise, the polynomial wouldn't have an extremum).
If $c<0$, you can see on the graph that the parabola  has a maximum obtained at a point with a negative abscissa. As this maximum is obtained at $x=-\frac d{2c}$ (result from high school), what can you conclude for $d$?
A: First, note $c$ and $d$ are fixed values. From the diagram,
$$f'(x) = (2c)x + d \lt 0 \tag{1}\label{eq1}$$
for all $x \ge 0$, as well as some $x \lt 0$ (in particular to just before the vertex at $x = \frac{-d}{2c}$). What you've shown is that $d \lt 6$ when checking at $x = 3$ with $c = -1$. However, $d$ must also satisfy the inequality \eqref{eq1} for all $x$ where $f'(x) \lt 0$, i.e., for $x \gt \frac{-d}{2c}$, with this including $x = 0$. As pointed out in another answer, at $x = 0$, you get $f'(0) = d$, so $d \lt 0$, regardless of the value of $c$. This is still consistent with your determination of $d \lt 6$ (with $c = -1$), but it's just more stringent.
A: 
If $f(x)=cx^2+dx+e$ for the function shown in the graph, then what values can $c$, $d$, and $e$ take on? 

Note: $c<0$, because the branches of the parabola open down. $e>0$, because $f(0)=e>0$. 

Since the slope of $f(x)$ is always negative, $f'(x)<0$ and since $f'(x)=(2c)x+d$, $(2c)x+d<0$.

No, the slope of $f(x)$ is not always negative, but:
$$\begin{cases}f'(x)>0, x<-\frac{d}{2c}\\ 
f'(x)<0,x>-\frac{d}{2c}\end{cases}$$
where $x=-\frac{d}{2c}$ is the $x$-coordinate of the vertex.

Since $c<0$, let's choose $c=−1$, and let's choose $x=3$. $(2)(−1)(3)+d<0$ which leads to $−6+d<0$ which leads to $d<6$. And the maximum value (in this case 6) of this open inequality can change based on what $c$ and $x$ values are chosen. So where am I wrong?

If $c=-1,x=3$, then $f'(x)<0\iff x>-\frac{d}{2c} \Rightarrow 3>\frac{d}{2} \iff d<6$ is necessary, but not sufficient condition for the given parabola. Why? Because, from the graph it is visible that the vertex is on the left of origin (i.e. $x<0$). So, the vertex must be $x=-\frac{d}{2c}=\frac{d}{2}<0 \Rightarrow d<0$, which is the sufficient condition. 
