# Write a cubic function that passes through the points. (-3, 0) (-1, 10) (0,0) (4, 0)Please explain your work, thank you! :)

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Let f(x) be the cubic function.

Then, we will write the function into the standard format:

f(x) = ax^3 + bx^2 + cx + d

Now to dteremine a,b,c,and d values, we will subsitute the points given.

The point (0,0):

==> f(0)=0+0++0+d= 0

==> d= 0

The point (4,0)

==> f(4) = a*4^3 + b*4^2 + c*4 = 0

==> 64a + 16b + 4c = 0

Divide by 4 to simplify:

==> 16a + 4b + c = 0.............(1)

Now the point (-3,0)

f(-3) = a*-3^3 + b*-3^2 + c*-3 = 0

==> -27a + 9b -3c = 0

divide by 3:

==> -9a + 3b -c = 0 ..............(2)

Now the point (-1,10)

f(-1) = a*-1^3 + b*-1^2 + c*-1 = 10

==> -a + b -c = 10...................(3)

Now we will solve the system:

using the elimination method, we will add (1) and (2):

==> 7a +7b = 0

divide by 7:

==> a+ b = 0...............(4)

==> 15a +5b = 10

Divide by 5 :

==> 3a + b = 2.............(5)

Now subtract (4) from (5)

==> 2a = 2

==> a= 1

==> b= -1

==> c= -12

==> f(x) = x^3 - x^2 -12x

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The curve that describes the cubic function passes through the given point, if and only if the coordinates of the points verify the expression of the function.

We'll write the cubic function:

f(x) = ax^3 + bx^2 + cx + d

The point (-3, 0) belongs to the graph of cubic function if and only if:

f(-3) = 0

f(-3) = a*(-3)^3 + b*(-3)^2 + c*(-3) + d

f(-3) = -27a + 9b - 3c + d

-27a + 9b - 3c + d = 0 (1)

The point  (-1, 10) belongs to the graph of cubic function if and only if:

f(-1) = 10

f(-1) = a*(-1)^3 + b*(-1)^2 + c*(-1) + d

f(-1) = -a + b - c + d

-a + b - c + d = 10 (2)

The point (0,0) belongs to the graph of cubic function if and only if:

f(0) = 0

d = 0

The point (4, 0) belongs to the graph of cubic function if and only if:

f(4) = 0

a*(4)^3 + b*(4)^2 + 4c + d = 0

64a + 16b + 4c = 0 (3)

We'll form the system from the equtaions (1),(2),(3):

-27a + 9b - 3c = 0

-a + b - c  = 10

64a + 16b + 4c = 0

We'll multiply (2) by (-3) and we'll add to (1):

3a - 3b + 3c - 27a + 9b - 3c = 30

We'll combine and eliminate like terms:

-24a + 6b = 30

We'll divide by 6:

-4a + b = 5 (4)

We'll multiply (2) by 4 and we'll add to (3):

-4a + 4b - 4c + 64a + 16b + 4c = 40

We'll combine and eliminate like terms:

60a + 20b = 40

We'll divide by 20:

3a + b = 2 (5)

We'll multiply (5) by (-1) and we'll add it to (4):

-3a - b - 4a + b = 5 - 2

-7a = 3

a = -3/7

We'll substitute a in (5):

-9/7 + b = 2

b = 2 + 9/7

b = 23/7

We'll substitute a and b in (2):

-a + b - c  = 10

3/7 + 23/7 - c = 10

26/7 - c = 10

c = 26/7 - 10

c = -44/7

The cubic function is:

f(x) = (-3/7)x^3 + (23/7)x^2 - (44/7)x

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Let y = ax^3+bx^2+cx+d  be the cubic function

Since the function is satisfied by these four points, (-3,0),(-1,10),(,0) and(4,0), we take the points one by one and determine the constants a,b,c,and d.

(0, 0) should satisfy:  a*0^3+b*0^2+c*0+d = 0. So d= =0.

Therefore the function is ax^3+bx^2+cx = y.

(-3,0) should satisfy : a(-3)^3+b*(-3)^2+x(3) = 0

We divide by (-3) and we get: 9a-3b+c = 0....(1).

(4,0) should satisfy: a*4^3+b*4^2+c*4 = 0.

We divide the equation by 4,and we get: 16a +4b+c = 0....(2).

(-1,10) should satisfy: a*(-1)^3+b(-1)^2+c*(-1) = 10. Or

-a+b-c = 10.....(3).

Eq(1)+Eq(3) :  8a-2b = 10. ...(4).

Eq (2)+eq(3):  15a+5b = 10. We divide this equation by 5 :

3a+b = 2....(5).

Eq(4)+2 eq(5) gives: 14a = 14. So a = 14/14 = 1.

Substitute a = 1 in (5): 3+b = 2. So b = 2- 3 = -1.

Using (1),  9a-3b+c = 0. Therefore c = 3b-9a = -3-9 = -12.

Therefore a = 1, b = -1 and c = 12.

Therefore the cubic function determined by the 4 points is x^3-x^2+12x = y.