a Determine the equation of the public cubic function with a

a) Determine the equation of the public cubic function with a zeros of 4 and -3, a turning point at (-3, 0) and passing through the point P(-2,2)

b) Write the equation of a cubic function, f(x), with zeros at -5, 0, and 2 and end behaviour f(x) x to positive infinity as x to negative infinity

Solution

( a) Let the equation of the cubic function be y = f (x ) = a( x – 4)(x + 3)( x – c) = ax³ – a (c +1)x² + a(c - 12)x +12a…..(1) , where a and c are constants.

Since the function passes through the point ( -2,2), on substituting x = -2 and y = 2 in the 1^st equation, we have 2 = a(-2)³ –a (c+1)(-2)² + a(c - 12)(-2) +12a or, 2 = -8a - 4a(c +1) - 2a(c – 12)+ 12a or, -8a – 4ac - 4a -2ac + 24a + 12a – 2 = 0 or, 24a - 6ac – 2 = 0 or, 3ac – 12a + 1 = 0…(2) Further, since the cubic function has a turning point at (-3,0), on substituting x = -3 and y = 0 in the 1^st equation, we have   a(-3)³ -a(c+1)(-3)² +a (c -12)(-3) + 12a = 0 or, -27a -9ac -9a -3ac + 36a + 12a = 0 or, 12a – 12ac = 0 or ac = a , so that c = 1….(3)   . On substituting c = 1 in the 2^nd equation, we have   3a – 12a + 1 = 0   or, 9a = 1    so that a = 1/9. Now, on substituting c = 1 and a = 1/9, in the equation of the function, we have y = f(x) = a( x – 4)(x + 3)( x – c) = 1/9(x -4)(x +3)(x -1).

( b) Let the equation of the required cubic function be y = f(x) = a x(x + 5)( x -2) where a is a constant. Since f(x) as x - , a has to be negative > Thus, the required equation is y = f(x) = a x(x + 5)( x -2) where a is negative constant.