Question

for the polynomial function f(x) = -2x⁴ - 6x³, answer the parts a through e.
a. use the leading coefficient test to determine the graph’s end behavior.
a. the graph of f(x) rises to the left and falls to the right.
b. the graph of f(x) falls to the left and rises to the right.
c. the graph of f(x) rises to the left and rises to the right.
d. the graph of f(x) falls to the left and falls to the right.
b. find the x-intercepts. state whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept.
the x-intercept(s) is/are
(type an integer or a decimal. use a comma to separate answers as needed. type each answer only once.)
at which x-intercept(s) does the graph cross the x-axis? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the x-intercept(s) at which the graph crosses the x-axis is/are
(type an integer or a decimal. use a comma to separate answers as needed. type each answer only once.)
b. there are no x-intercepts at which the graph crosses the x-axis.
at which x-intercept(s) does the graph touch the x-axis and turn around? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the x-intercept(s) at which the graph touches the x-axis and turns around is/are
(type an integer or a decimal. use a comma to separate answers as needed. type each answer only once.)
b. there are no x-intercepts at which the graph touches the x-axis and turns around.
c. find the y-intercept.
the y-intercept is

Explanation:

Step1: Analyze Leading Coefficient and Degree

The polynomial is \( f(x) = -2x^4 - 6x^3 \). The leading term is \( -2x^4 \), so the leading coefficient is \(-2\) (negative) and the degree is \(4\) (even). For end - behavior: when the degree is even and the leading coefficient is negative, as \( x
ightarrow+\infty \), \( f(x)
ightarrow-\infty \) (falls), and as \( x
ightarrow-\infty \), \( f(x)
ightarrow-\infty \) (falls). So the graph falls to the left and falls to the right, which is option D.

Step2: Find x - intercepts

To find the x - intercepts, set \( f(x)=0 \), so \( -2x^4 - 6x^3=0 \). Factor out \(-2x^3\): \( -2x^3(x + 3)=0 \). Then we have two equations: \(-2x^3 = 0\) or \(x + 3=0\). Solving \(-2x^3=0\) gives \(x = 0\) (with multiplicity 3), and solving \(x + 3=0\) gives \(x=-3\) (with multiplicity 1).

Step3: Determine x - intercept behavior

• For \(x=-3\) (multiplicity 1, odd), the graph crosses the x - axis.
• For \(x = 0\) (multiplicity 3, odd? Wait, multiplicity 3 is odd? Wait, no, multiplicity 3 is odd. Wait, but let's re - check. Wait, the factor is \(x^3\), so multiplicity 3. When multiplicity is odd, the graph crosses the x - axis? Wait, no, if multiplicity is odd, the graph crosses the axis; if even, it touches and turns. Wait, \(x = 0\) has multiplicity 3 (odd), so the graph crosses at \(x = 0\)? Wait, no, wait the original factoring: \(f(x)=-2x^3(x + 3)\). So the roots are \(x = 0\) (multiplicity 3) and \(x=-3\) (multiplicity 1).
• For the x - intercepts where the graph crosses: \(x=-3,0\) (since both multiplicities are odd).
• For the x - intercepts where the graph touches and turns: Since all multiplicities are odd, there are no x - intercepts where the graph touches and turns.

Step4: Find y - intercept

To find the y - intercept, set \(x = 0\) in \(f(x)\). Then \(f(0)=-2(0)^4-6(0)^3 = 0\).

Answer:

a. D. The graph of \(f(x)\) falls to the left and falls to the right.
b. The x - intercepts are \(-3,0\). At \(x=-3\) and \(x = 0\) the graph crosses the x - axis (since their multiplicities are odd), and there are no x - intercepts where the graph touches and turns.
c. The y - intercept is \(0\).