To find the x-intercepts of the function $f(x) = \frac{5x^3 - 9x}{4x^2 - 3x + 2}$, we need to determine where the function equals zero. An x-intercept occurs when $f(x) = 0$, which means the numerator of the function must be zero (since a fraction is zero only if its numerator is zero).

Thus, we solve for $x$ by setting the numerator equal to zero:

$5x^3 - 9x = 0$

Factor the equation:

$x(5x^2 - 9) = 0$

This gives two factors:

1. $x = 0$
2. $5x^2 - 9 = 0$

Solve $5x^2 - 9 = 0$ by adding 9 to both sides:

$5x^2 = 9$

Now, divide by 5:

$x^2 = \frac{9}{5}$

Take the square root of both sides:

$x = \pm \sqrt{\frac{9}{5}} = \pm \frac{3}{\sqrt{5}} \approx \pm 1.525$

So, the values of $x$ that make the numerator zero are:

$x = 0, \quad x \approx -1.525, \quad x \approx 1.525$

Now we check which of these solutions match the options provided:

• The values $x \approx -1.525$, $x \approx 0.525$, and $x = 2$ appear in option C.

Therefore, the correct answer is:

C) $x = -1.525$, $x = 0.525$, and $x = 2$ only.

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Would you like more details on any of the steps or have further questions?

Here are 5 related questions to explore:

1. How do you find vertical asymptotes of a rational function?
2. What are critical points and how do you find them?
3. How do you find the local maxima and minima of a function?
4. What is the significance of the derivative in finding x-intercepts?
5. How do you determine horizontal asymptotes of a rational function?

Tip: Always check both the numerator and the denominator when finding intercepts and asymptotes for rational functions.