Question 1038159: 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?
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! ax^3+bx^2+cx+d is the polynomial
-a+b-c+d=15
d=0, so one intercept is (0,0). The reason d=0 is that f(0)=0. When f(0) is applied to the function, all the x terms disappear, since they are all zero. The only term left is d, the constant. But d is 0, because f(0)=0.
a+b+c=-5
8a+4b+2c=12
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-a+b+c=15. This comes from the first equation, removing d, since it is 0. I can then add that and the one below it.
a+b+c=-5
2b+2c=10
b+c=5, dividing by 2.
a=-10
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-80+10+2b=12, because 2b+2c+10
2b=82
b=41
c=-36
-10x^3+41x^2-36x=0
factor out an -x
-x(10x^2-41x+36)=0
x=0
x=(1/20)(41 +/-sqrt (1681-1440); sqrt 241=15.52
x=56.52/20, or 2.83
x=25.48/20, or 1.274
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 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?
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Using the form ax^3+bx^2+cx+d = y,
substitute each of the x/y pairs
to create 4 equations in a,b,c,d.
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Example::
Using f(-1)=15 you get -a+b-c+d = 15
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Solve the 4-equation system for a,b,c,d
to find the cubic you want.
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Cheers,
Stan H.
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