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The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

 Feb 19, 2019
 #1
avatar+5038 
+2

\(f(x) = a x^3 + b x^2 + c x + d\\ f(0)= 0 \Rightarrow d = 0\\ \text{using the rest of the values we obtain 3 equations}\\ a + b + c =-5\\ -a + b -c =15\\ 8a+4b + 2c = 12\)

 

\(\text{adding the first two}\\ 2b = 10,~b = 5\)

 

\(a+c=-10\\ 8a+2c = -8\\ \text{subtract twice eq 1 from eq 2}\\ 6a = 12,~a=2\\ c= -12\)

 

\(f(x) = 2x^3 + 5x^2 -12x\)

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 Feb 19, 2019
 #2
avatar+100439 
+2

We have the form

 

ax^3 + bx^2 + cx + d

 

If f(0) = 0.....then d = 0       and we can solve this system

 

a(-1)^3 + b(-1)^2 + c(-1)  =  15

a(1) ^3 + b(1)^2 + c(1) = - 5

a(2)^3 + b(2)^2 + c(2) = 12             simplify

 

 

-a + b - c   =     15            (1)

a + b + c  =     -5             (2)

8a + 4b + 2c = 12       ⇒ 4a + 2b + c = 6      (3)

 

Add (1) and (2)

2b = 10

b = 5

 

Add (1) and (3)

3a + 3b = 21

a + b = 7

 a + 5 = 7

a = 2

 

And using (1) to find c  we have

-2 + 5 - c = 15

3 - c = 15

3 - 15 = c

-12 = c

 

So...the polynomial is

 

2x^3 + 5x^2 - 12x 

 

To find the x intercepts (roots) we have

 

2x^3 + 5x^2 - 12x = 0         factor

 

x (2x^2 + 5x^2 - 12) = 0

 

x ( 2x  - 3) (x + 4) =  0

 

Setting each factor to 0  and solving for x produces the x intercepts of

 

x = 0       x = 3/2       and      x =  -4

 

cool cool cool

 Feb 19, 2019

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