Finding the equation of a function

Dec 2008
509
2
Hi

I need help on the following question:

1) The graph of a polynomial function shows x-axis intercepts at x = -1 and x = 3 and touches the x-axis (with a turning-point) at x = 2. The y-axis intercepts is at y = 24. Assuming the polynomial has degree four, write its equation in factorized form.

The equation i got is \(\displaystyle y = (x+1)(x-3)(x-2)^2.\)
This is incorrect, how did the book's answer get \(\displaystyle y = -2(x+1)(x-3)(x-2)^2\)

P.S
 
Sep 2008
1,261
539
West Malaysia
Hi

I need help on the following question:

1) The graph of a polynomial function shows x-axis intercepts at x = -1 and x = 3 and touches the x-axis (with a turning-point) at x = 2. The y-axis intercepts is at y = 24. Assuming the polynomial has degree four, write its equation in factorized form.

The equation i got is \(\displaystyle y = (x+1)(x-3)(x-2)^2.\)
This is incorrect, how did the book's answer get \(\displaystyle y = -2(x+1)(x-3)(x-2)^2\)

P.S
hi

the question also assumes that the polynomial has integer coefficient .

so f(x)=a(x+1)(x-3)(x-2)^2

f(0)=24

24=a(0+1)(0-3)(0-2)^2

a=24/(-12)=-2

hence , f(x)=-2(x+1)(x-3)(x-2)^2 as suggested by the answer
 
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