Asked by moe
2. The following 6 points belong to a certain function: (1, –4), (2, –16), (–2,–4), (0, 2), (–1,
2), (3,–34). About this function, please answer two questions:
a. What type of function does this set of points produce? Justify your answer. [2]
b. Write an equation for this function. [5]
2), (3,–34). About this function, please answer two questions:
a. What type of function does this set of points produce? Justify your answer. [2]
b. Write an equation for this function. [5]
Answers
Answered by
oobleck
Maybe it would be easier to arrange them left to right.
Of course, if you just plot them, it should be clear.
(-2,-4) (-1,2) (0,2) (1,-4) (2,-16) (3,-34)
It looks like a parabola with axis of symmetry at x = -1/2
To be sure it's quadratic, check the differences.
6,0,-6,-12,-18
-6 -6 -6 -6
Since the 2nd differences are constant, it is a quadratic, and we know that its equation in vertex form is
y = a(x + 1/2)^2 + k
or,
y = a(2x+1)^2 + k
using (1,-4) and (0,2), we have
a+k = 2
9a+k = -4
so a = -3/4
k = 11/4
and
y = -3/4 (2x+1)^2 + 11/4 = -3x^2-3x+2
Of course, if you just plot them, it should be clear.
(-2,-4) (-1,2) (0,2) (1,-4) (2,-16) (3,-34)
It looks like a parabola with axis of symmetry at x = -1/2
To be sure it's quadratic, check the differences.
6,0,-6,-12,-18
-6 -6 -6 -6
Since the 2nd differences are constant, it is a quadratic, and we know that its equation in vertex form is
y = a(x + 1/2)^2 + k
or,
y = a(2x+1)^2 + k
using (1,-4) and (0,2), we have
a+k = 2
9a+k = -4
so a = -3/4
k = 11/4
and
y = -3/4 (2x+1)^2 + 11/4 = -3x^2-3x+2
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