x intercepts of f(x)= 2x-4/3x^2-7x+2
​


To find the \(x\)-intercepts of a function, we set \(f(x) = 0\) and solve for \(x\). 

Given the function: 
\[
f(x) = \frac{2x - 4}{3x^2 - 7x + 2}
\]

We set the numerator equal to zero, as this will give the \(x\)-intercepts (because a fraction is equal to zero when the numerator is zero, provided the denominator is not zero):

\[
2x - 4 = 0
\]

Solving for \(x\):
\[
2x = 4
\]
\[
x = 2
\]

Thus, the \(x\)-intercept is at \(x = 2\), but we need to check that the denominator is not zero at this point. Let's substitute \(x = 2\) into the denominator:

\[
3(2)^2 - 7(2) + 2 = 3(4) - 14 + 2 = 12 - 14 + 2 = 0
\]

Since the denominator is zero at \(x = 2\), this means the function is undefined at this point, so there is no \(x\)-intercept.

### Conclusion:
There are no \(x\)-intercepts because \(x = 2\) causes the denominator to be zero, resulting in an undefined value at that point.

Would you like further details or have any other questions?

Here are some related questions you may find interesting:
1. How do we find vertical asymptotes for rational functions?
2. What is the significance of a function being undefined at a particular point?
3. How can we find the domain of a rational function?
4. What happens at points where both the numerator and denominator are zero?
5. How do we check for horizontal asymptotes in a rational function?

**Tip:** Always check if the denominator is zero when finding \(x\)-intercepts for rational functions.

Explore how to find the x-intercepts of the rational function f(x) = (2x - 4) / (3x^2 - 7x + 2). This problem highlights key algebraic concepts, including finding intercepts and checking undefined points where the denominator equals zero. Suitable for Grades 9-12.

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