The graph of a polynomial function has the following characteristics:
• Its domain and range are the set of all real numbers.
• There are turning points at and -2, 0, 3.
• It must be a quartic function.
a) Draw the graphs of two different polynomial functions that have
these three characteristics.
b) What additional characteristics would ensure that only one graph
could be drawn?
How can I create a graph with these guidelines given? Can you outline your steps on how to make the equasion and graph?
Edit: I don't want the actual graph, jut how to make one with the given instructions.
Answer: f(x) = 1/4x^4 - 1/3x^3 -3x^2
f(x) = 1/4x^4 - 1/3x^3 - 3x^2 - 1
• Its domain and range are the set of all real numbers.
• There are turning points at and -2, 0, 3.
• It must be a quartic function.
a) Draw the graphs of two different polynomial functions that have
these three characteristics.
b) What additional characteristics would ensure that only one graph
could be drawn?
How can I create a graph with these guidelines given? Can you outline your steps on how to make the equasion and graph?
Edit: I don't want the actual graph, jut how to make one with the given instructions.
Answer: f(x) = 1/4x^4 - 1/3x^3 -3x^2
f(x) = 1/4x^4 - 1/3x^3 - 3x^2 - 1
-
Let f be such a function. Then its derivative f' must vanish at -2, 0, and 3, so it is divisible by x(x+2)(x-3). Since f is quartic, f' must be cubic, so we must have
f'(x) = cx(x+2)(x-3)
for some constant c. Conversely, it is easy to see that any f with f'(x) in this form and c =/= 0 will satisfy the given requirements. Multiplying this out yields
f''(x) = c x^3 - cx^2 - 6cx.
Integrating, we find
f(x) = (c/4) x^4 - (c/3) x^3 - 3cx^2 + C
for some constant C. In other words, the solutions are any functions in the above form for any real number C and any nonzero real number c. The given examples correspond to c=1 and C = 0, 1.
There are lots of way to ensure that there is only one possible answer. For example. you can specify f(0) and f'(1), Then there is only one possible function, because C=f(0) and -6c=f'(1).
f'(x) = cx(x+2)(x-3)
for some constant c. Conversely, it is easy to see that any f with f'(x) in this form and c =/= 0 will satisfy the given requirements. Multiplying this out yields
f''(x) = c x^3 - cx^2 - 6cx.
Integrating, we find
f(x) = (c/4) x^4 - (c/3) x^3 - 3cx^2 + C
for some constant C. In other words, the solutions are any functions in the above form for any real number C and any nonzero real number c. The given examples correspond to c=1 and C = 0, 1.
There are lots of way to ensure that there is only one possible answer. For example. you can specify f(0) and f'(1), Then there is only one possible function, because C=f(0) and -6c=f'(1).