Which of the following can also be written as a recursive sequence?
2 answers:
You might be interested in
Answer:
The solutions are:
Step-by-step explanation:
We have the following quadratic equation
We can rewrite the equation as follows
Now we use the quadratic formula to solve the equation
For an equation of the form the quadratic formula is:
In this case:
Then:
Your calculator's cubic regression function can tell you the equation is
... f(x) = 2x³ + 5x² -12x = x(x +4)(2x-3)
The x-intercepts are -4, 0, +1.5.
_____
If you want to solve this "by hand", you can first of all recognize that since there is an x-intercept at 0, the cubic will only have three coefficients. That is, you can write the equation as
... f(x) = ax³ + bx² + cx
Substituting the given points (except (0, 0)) gives three linear equations in a, b, c.
... -a +b -c = 15 . . . . . for x=-1
... a + b + c = -5 . . . . for x=1
... 8a +4b +2c = 12 . . for x=2
adding the first two equations gives 2b=10, or b=5. Now, you can reduce the system to
... a + c = -10
... 4a +c = -4
Subtracting the first of these equations from the second gives 3a=6, or a=2. That tells you c=-12 (from a+c=10).
Then your equation is
... f(x) = x(2x² +5x -12)
Factoring by any of the usual techniques, or graphing, or using the quadratic formula will tell you the zeros (x-intercepts) are as above.
_____
Since the input values are sequential, you can also develop the function from differences of the output values. Those are 15, 0, -5, 12. First differences are -15, -5, +17. Second differences are +10, +22. The third difference is 12. Using the first of these differences in appropriate places in the interpolating polynomial formula, we have
... f(x) = 15 + (x+1)(-15 + (x)/2·(10 + (x-1)/3·(12))) = 2x³ +5x² -12x . . . . as above
