The problem asks to determine the real zeros, their multiplicities, and to sketch the graph of the given polynomial function:

$f(x) = 2x^4 + x^3 - 6x^2$

Here are the steps to solve it:

1. Find the real zeros of the polynomial:

Factor the polynomial to find its zeros.

$f(x) = 2x^4 + x^3 - 6x^2 = x^2(2x^2 + x - 6)$

Now, factor the quadratic part $2x^2 + x - 6$:

$2x^2 + x - 6 = (2x - 3)(x + 2)$

Thus, the factored form of $f(x)$ is:

$f(x) = x^2(2x - 3)(x + 2)$

2. Determine the real zeros:

From the factored form $f(x) = x^2(2x - 3)(x + 2)$, the real zeros are:

• $x = 0$ (with multiplicity 2, since $x^2$ gives a double root),
• $x = \frac{3}{2}$ (with multiplicity 1),
• $x = -2$ (with multiplicity 1).

3. Check the graphs provided:

We now need to match this to the correct graph by looking for:

• A double root at $x = 0$ (which means the graph will touch the x-axis at $x = 0$ but not cross it),
• Single roots at $x = \frac{3}{2}$ and $x = -2$ (the graph will cross the x-axis at these points).

The graph that matches these criteria is Option A. It shows a touch at $x = 0$ (indicating multiplicity 2) and crossings at $x = \frac{3}{2}$ and $x = -2$.

Answer:

• Real Zeros: $0$ (multiplicity 2), $\frac{3}{2}$ (multiplicity 1), $-2$ (multiplicity 1).
• Correct Graph: Option A.

Would you like further details or explanations on any part of this solution?

Here are 5 related questions you might want to explore:

1. How does the multiplicity of a zero affect the shape of the graph at that point?
2. What are the steps for factoring higher-degree polynomials?
3. Can a polynomial of degree 4 have more than 4 real zeros?
4. How can you determine end behavior of a polynomial from its leading term?
5. How does the sign of the leading coefficient affect the graph's direction?

Tip: When a polynomial has a zero with an even multiplicity, the graph "bounces" off the x-axis at that zero instead of crossing it.