Find the coefficients $a, b, c$ and $d$ so that the curve shown in the accompanying figure is the graph of the equation.

Question

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?

Answer 1

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 $d=10$.

Answer 2

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 $c$.

Answer 3

$\begin{vmatrix} x^3&x^2&x&1&y\\ 0^3&0^2&0&1&10\\ 1^3&1^2&1&1&7\\ 3^3&3^2&3&1&-11\\ 4^3&4^2&4&1&-14\\ \end{vmatrix}= \begin{vmatrix} x^3&x^2&x&1&y\\ 0&0&0&1&10\\ 1&1&1&1&7\\ 27&9&3&1&-11\\ 64&16&4&1&-14\\ \end{vmatrix}= -72x^3+432x^2-144x-720+72y= x^3-6x^2+2x+10-y=0$