Write the cubic function whose graph is shown with that points passing through (3,0) (0,18) (-1,0) (-3,0)

Question

f(x)=

Answer 1

The general form of a cubic equation is f(x) = Ax³+Bx²+Cx+D

You have 4 points so we can get 4 equations and solve the system of equations.

From (-3, 0):

f(-3) = A(-3)³ + B(-3)² + C(-3) + D = 0 ⇒ -27A + 9B - 3C + D = 0

(-1,0)

f(-1) = A(-1)³ + B(-1)² + C(-1) + D = 0 ⇒ -A + B - C + D = 0

(0, 18)

f(0) = D = 18 (so we have our first coefficient, D = 18)

(3,0)

f(3) = A(3)³ + B(3)² + C(3) + D = 0 ⇒ 27A + 9B + 3C + D = 0

Rewriting these and plugging in for D = 18, then reducing we get:

9A - 3B + C = 18

A - B + C = 18

9A + 3B + C = -6

Now solve the system of linear equations (I used matrices, but difficult to reproduce here).

I came up with A = -2, B = -2, C = 18, and D = 18 hence

f(x) = -2x³ - 2x² + 18x + 18

Answer 2

Hi Cole,

Cubic functions take the form y=a(x-b)(x-c)(x-d). b, c and d are the x-intercepts of this function. From the graph we see that -3,-1 and 3 are the the x-intercepts. We can insert them into the function -

y = a(x-(-3))(x-(-1))(x-3))

This simplifies to

y = a(x+3)(x+1)(x-3)

Now we just have to find a. To do this we'll use the remaining point - (0,18). Because we know that at x=0 y=18, we can plug those values into the equation to solve for a.

y = a(x+3)(x+1)(x-3)

18=a(0+3)(0+1)(0-3)

18=a(3)(1)(-3)

18= -9a

a= -2

Now plug a into the final equation -

y = -2(x+3)(x+1)(x-3)