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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?

 

if someone could help me out that'dbe great, thanks!

 Aug 7, 2019
 #1
avatar+103693 
+1

Let me ask YOU some questions.  Just to help you learn.

 

One of them can be seen immediately.  What one is that?

 

How many (at most) can there be?

 Aug 7, 2019
 #2
avatar+103122 
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A 3rd degree polynomial has the  form      ax^3  + bx^2 + cx + d

 

Since  f(0)  = 0     then   x = 0   is one of the y intercepts  and    d  = 0

 

So....we can solve  for  a, b and  c     as follows :

 

a(-1)^3  + b(-1)^2  + c(-1) =  15  ⇒      -a   +  b - c  = 15     (1)

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

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

 

Adding  (1)  and (2)   we have that     2b  =  10 ⇒    b  = 5

Sub this into  each equation for b and we  have that

 

-a +5 - c  =  15 ⇒    -a  - c =  10    (4)

a +5  + c  = -5  ⇒   a  + c  = -10        (5)

8a + 4(5) + 2c =  12  ⇒   8a  + 2c = -8   (6)

 

Multiply (5)  by -2  and add it to (6)   and we get that     6a  =12 ⇒   a  = 2

Using (5)  to find c  we have that    2 + c  = -10 ⇒    c  =-12

 

So....the polynomial is   f(x)  =   2x^3 + 5x^2  - 12x      set to 0 and factor and we  get  that

 

(2x  - 3) (x + 4 )  = 0

 

Setting each factor to  0 and  solving  for  x  and  we  get the  other  two x intercepts  of   x  = 3/2    and x =  -4

 

See the graph here  : https://www.desmos.com/calculator/ef2xuaesrm

 

 

 

cool cool cool

 Aug 7, 2019

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