Question

A cubic function of the y=ax^3+bx^2+cx+d yields the following table:

Answer 1

Answer:

a=-5

b=4

c=-1

d=0

Step-by-step explanation:

The given cubic equation is of the form;

$y = a {x}^{3} + b {x}^{2} + cx + d$

When x=0 and y=0, we get:

$0= a ({0})^{3} + b( {0})^{2} + c(0) + d$

$d = 0$

when x=1 and y=-2, we get:

$- 2= a ({1})^{3} + b( {1})^{2} + c(1) \\ a + b + c = - 2$

When x=2 and y=-26, we get:

$- 26= a ({2})^{3} + b( {2})^{2} + c(2) + d \\ - 26 = 8a + 4b + 2c \\ 4a + 2b + c = - 13$

When x=3 and y=-54, we get:

$- 54= a ({3})^{3} + b( {3})^{2} + c(3) + d \\ - 54 = 27a + 9b + 3c \\ 9a + 3b + c = - 18$

We need to solve the system:

$a + b + c = - 2 \\ 4a + 2b + c = - 13 \\ 9a + 3b + c = - 18$

From the top equation:

$c = - 2 - a - b$

Put this expression into equation 2 and 3

$4a + 2b - 2 - a - b = - 13 \\ 3a + b = - 11$

also

$9a + 3b - 2 - a - b = - 18 \\ 8a + 2b = - 16 \\ 4a + b = - 8$

Now we solve the system:

$3a + b = - 11 \\ 4a + b = - 16$

subtract the top equation from the down one to get:

$4a - 3a = - 16 - - 11 \\ a = - 5$

This implies that:

$3( - 5) + b = - 11 \\ - 15 + b = - 11 \\ b = - 11 + 15 = 4$

and

$c = - 2 - - 5 - 4 = - 1$