Find the coefﬁcients $a, b, c$ and $d$ so that the curve shown in the accompanying ﬁgure is the graph of the equation. 
Find the coefﬁcients $a, b, c$ and $d$ so that the curve shown in the accompanying ﬁgure is the graph of the equation $y = ax^3 + bx^2 + cx + d$. 
I have no clue how to solve this. This looks like nothing from any example in my book. How do I solve this? Where do I even start?
 A: Use the general form of your equation: 
$$y=ax^3+bx^2+cx+d$$
Substitute known pairs of $(x,y)$. You have four unknowns and you will get 4 equations for: $(0,10)$, $(1,7)$, $(3,−11)$, and $(4,−14)$. They are:
$$10 = d$$
$$7 = 1a+1b+1c+d $$
$$-11 = 27a+9b+3c+d $$
$$-14 = 64a+16b+4c+d $$
Then it's just simple linear algebra or CAS usage to find $a$, $b$, $c$, $d$. 
You should get: $a=1$, $b=−6$, $c=2$, and $d=10$.
A: In the book, they give you all required information. As already said by jojek, you have a function $$y=ax^3+bx^2+cx+d$$ which must go through the points $(0,10)$,$(1,7)$,$(3,-11)$,$(4,-14)$. So just write these conditions accordingly, that is to say $$10=d$$ $$7=a+b+c+d$$ $$-11=27a+9b+3c+d$$  $$-14=64a+16b+4c+d$$ from which you must extract the values of $a,b,c,d$.
I am sure that you can take from here.
Added later to this answer
The first equation gives $d=10$. So putting this value in the other equations, we have $$a+b+c=-3$$ $$27a+9b+3c=-21$$ $$64a+16b+4c=-24$$ whcih simplify to $$a+b+c=-3$$ $$9a+3b+c=-7$$ $$16a+4b+c=-6$$ Now, eliminate $c$ which gives $c=-3-a-b$; replace and continue the same way until you have a single equation in $a$; solve it and go backwards for getting $b$, then $c$.
A: $\begin{vmatrix}
x^3&x^2&x&1&y\\
0^3&0^2&0&1&10\\
1^3&1^2&1&1&7\\
3^3&3^2&3&1&-11\\
4^3&4^2&4&1&-14\\
\end{vmatrix}=
\begin{vmatrix}
x^3&x^2&x&1&y\\
0&0&0&1&10\\
1&1&1&1&7\\
27&9&3&1&-11\\
64&16&4&1&-14\\
\end{vmatrix}=
-72x^3+432x^2-144x-720+72y=
x^3-6x^2+2x+10-y=0$
