Let's tackle the questions step-by-step.
Finding the Zeros of a Polynomial Function:
Question 1:
A)
The polynomial function provided is \( h(x) = -3x^2(5x + 10)(x - 2) \). To fill in the blanks with whole numbers, let's set \( 5x + 10 = 5(x + 2) \).
Thus, the polynomial is: \[ h(x) = -3x^2(5(x + 2))(x - 2) \]
B)
To find the zeros of \( h(x) \), we set \( h(x) = 0 \): \[ -3x^2(5(x + 2))(x - 2) = 0 \]
This equation will be zero if any of the factors are zero:
-
\( -3x^2 = 0 \)
- This gives \( x = 0 \).
-
\( 5(x + 2) = 0 \)
- This gives \( x + 2 = 0 \) or \( x = -2 \).
-
\( (x - 2) = 0 \)
- This gives \( x = 2 \).
The zeros of \( h(x) \) are \( x = 0, -2, 2 \).
Question 2:
A)
For \( j(x) = \frac{12x^3}{5} - \frac{27x}{5} \), factor out the common term: \[ j(x) = \frac{1}{5}(12x^3 - 27x) = \frac{1}{5}(3x(4x^2 - 9)) = \frac{1}{5}(3x(2x - 3)(2x + 3)) \]
Set \( j(x) = 0 \):
- \( 3x = 0 \) → \( x = 0 \)
- \( 2x - 3 = 0 \) → \( x = \frac{3}{2} \)
- \( 2x + 3 = 0 \) → \( x = -\frac{3}{2} \)
The zeros of \( j(x) \) are \( x = 0, \frac{3}{2}, -\frac{3}{2} \).
Sketching Between the Zeros of a Polynomial Function:
Question 1 (for \( h(x) \)):
A)
The zeros found are \( -2, 0, 2 \). The intervals to test are:
-
\( (-\infty, -2) \)
- Choose \( x = -3 \) → \( h(-3) = -3(-3)^2(5(-3 + 2))(-3 - 2) = - \) (negative).
-
\( (-2, 0) \)
- Choose \( x = -1 \) → \( h(-1) = -3(-1)^2(5(-1 + 2))(-1 - 2) = + \) (positive).
-
\( (0, 2) \)
- Choose \( x = 1 \) → \( h(1) = -3(1)^2(5(1 + 2))(1 - 2) = - \) (negative).
-
\( (2, \infty) \)
- Choose \( x = 3 \) → \( h(3) = -3(3)^2(5(3 + 2))(3 - 2) = - \) (negative).
So, the behavior is:
- Above 0 (positive) for \( (-2, 0) \)
- Below 0 (negative) for other intervals.
B)
On Graph 1, plot points corresponding to the zeros and the behavior found above.
Question 2 (for \( j(x) \)):
A)
The zeros found are \( -\frac{3}{2}, 0, \frac{3}{2} \). The intervals to test are:
-
\( (-\infty, -\frac{3}{2}) \)
- Choose \( x = -2 \) → \( j(-2) = \frac{1}{5}(12(-2)^3 - 27(-2)) < 0 \) (negative).
-
\( (-\frac{3}{2}, 0) \)
- Choose \( x = -1 \) → \( j(-1) = \frac{1}{5}(12(-1)^3 - 27(-1)) > 0 \) (positive).
-
\( (0, \frac{3}{2}) \)
- Choose \( x = 1 \) → \( j(1) = \frac{1}{5}(12(1)^3 - 27(1)) < 0 \) (negative).
-
\( (\frac{3}{2}, \infty) \)
- Choose \( x = 2 \) → \( j(2) = \frac{1}{5}(12(2)^3 - 27(2)) > 0 \) (positive).
So, the behavior is:
- Above 0 (positive) for \( (-\frac{3}{2}, 0) \) and \( (x > \frac{3}{2}) \).
End Behavior of Polynomial Functions:
Question 1 (for \( h(x) \)):
A)
The leading term in \( h(x) = -3x^2(5)(x - 2) \) is \( -15x^3 \).
- As \( x \to \infty \), \( h(x) \to -\infty \).
- As \( x \to -\infty \), \( h(x) \to \infty \).
B)
On Graph 1, incorporate the end behavior by extending the lines to reflect this.
Question 2 (for \( j(x) \)):
A)
The leading term is \( \frac{12}{5}x^3 \).
- As \( x \to \infty \), \( j(x) \to \infty \).
- As \( x \to -\infty \), \( j(x) \to -\infty \).
B)
On Graph 2, incorporate the end behavior by extending the lines accordingly.
Note:
- For both graphing responses, it's essential to accurately represent points and curves to capture the polynomial's behavior visually.
- Ensure that all work is clearly noted with labels for clarity when presenting.
This should guide you through completing the assignment effectively!
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