Question

f(x) = -x³-8;
f(x) = -x³ + 12x.​

Answer 1

To find the x-intercepts of each function, we set f(x) equal to zero and solve for x:

For f(x) = -x³-8:
0 = -x³-8
x³ = -8
x = -2 (since (-2)³ = -8)

Therefore, the x-intercept of f(x) = -x³-8 is -2.

For f(x) = -x³ + 12x:
0 = -x³ + 12x
0 = x(-x² + 12)
x = 0 or x = ±√12 (using the quadratic formula to solve for the other two roots)

Therefore, the x-intercepts of f(x) = -x³ + 12x are 0 and ±√12.

To find the y-intercepts of each function, we set x equal to zero and evaluate f(0):

For f(x) = -x³-8:
f(0) = -(0)³-8 = -8

Therefore, the y-intercept of f(x) = -x³-8 is -8.

For f(x) = -x³ + 12x:
f(0) = -(0)³ + 12(0) = 0

Therefore, the y-intercept of f(x) = -x³ + 12x is 0.

To determine the symmetry of each function, we can check whether the function is even or odd:

For f(x) = -x³-8:
f(-x) = -(-x)³-8 = -(-x³)-8 = -(-x³ + 8) = x³ - 8
Since f(-x) ≠ -f(x), the function is not even.
Since f(-x) ≠ f(x), the function is not odd.
Therefore, the function does not have any symmetry.

For f(x) = -x³ + 12x:
f(-x) = -(-x)³ + 12(-x) = x³ - 12x
Since f(-x) = -f(x), the function is odd.
Therefore, the function has rotational symmetry about the origin.

Finally, we can sketch the graphs of each function using the intercepts and symmetry information:

For f(x) = -x³-8, we know that the graph passes through (-2, 0) and (-8, 0) and has no symmetry.

For f(x) = -x³ + 12x, we know that the graph passes through (0, 0), (0, √12), and (0, -√12) and has rotational symmetry about the origin.

Here is a rough sketch of the graphs of both functions:

```
|
0| + . +
| *
| * *
| * *
-10 + * * * * * * * *
| * *
| * *
| * *
-20 +-------------------------->
-4 -2 0 2 4 6 8
```