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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

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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\$?

Guest Apr 27, 2017
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The polynomial will have the form

P(x)  = Ax^3  + Bx^2  + Cx + D

And since (0, 0)  is in the graph   then

A(0)^3  + B(0)^2  + C(0) + D  = 0      so...... D  = 0

And we have these three equations

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  →  4A + 2B + C  = 6   (3)

Adding (1)  and (2)   we have that 2B   =  10      →  B  = 5

Using (2)  and (3)  and subbing for B we have

A + 5 + C  =  -5   →   A + C  = -10    →   C  = -10  - A     (4)

4A + 10 + C  =  6  →   4A + C = -4  →  C  = -4 - 4A  (5)

Set  (4)  and (5)  equal

-10 - A   =  -4 - 4A

3A  = 6

A  =  2

Using (4)

C = -10 -  (2)

C  = -12

So.....the polynomial is

P(x)   =  2x^3  + 5x^2  -12x

The x intercepts can be found when P(x)  = 0

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 the x intercepts are    -4,  0 and  3/2

CPhill  Apr 27, 2017