Answer:
- f(x) = x^3 -4x
- f(x) = -2x^3 +8x
Step-by-step explanation:
The zeros are at -2, 0, and +2, so the function will be of the form ...
f(x) = k·(x +2)(x)(x -2) . . . . . for some vertical expansion factor k
You recognize that (x+2)(x-2) is the factoring of the difference of squares, so ...
f(x) = k·x·(x^2 -4)
If we let x=1, we get
f(1) = k·(1)(1 -4) = -3k
For the first graph, it looks like we have ...
f(1) = -3 = -3k . . . . so, k = 1
For the second graph, it looks like we have ...
f(1) = 6 = -3k . . . . so, k = -2
_____
In standard form, the first graph is described by ...
f(x) = 1·x·(x^2 -4) = x^3 -4x
The second graph is described by ...
f(x) = -2x·(x^2 -4) = -2x^3 +8x
Answer:
Not valid
Step-by-step explanation:
Because 12 is minor than 15
80000000000000000p0000000000
Let’s see! Substituting 3 for x and -1 for y:
(3 + 1)^2 + (-1 - 1)^2 = 16
4^2 + (-2)^2 = 16
16 + 4 = 16
20 = 16
20 is clearly not equal to 16, so we know the point (3, -1) can’t lie on that circle.
İf you find problem on answer,respond