f(x) = (2x^2+7x+5)/(3x^2-1x-10); This function has:

1) A y-intercept at the point 
  

2) x-intercepts at the point(s) 
  

3) Vertical asymptotes at x = 


Let's analyze the given function step by step to find:

\[
f(x) = \frac{2x^2 + 7x + 5}{3x^2 - x - 10}
\]

### 1) Y-intercept:
To find the y-intercept, substitute \( x = 0 \) into the function and simplify:

\[
f(0) = \frac{2(0)^2 + 7(0) + 5}{3(0)^2 - (0) - 10} = \frac{5}{-10} = -\frac{1}{2}
\]

Thus, the y-intercept is at the point \( (0, -\frac{1}{2}) \).

### 2) X-intercepts:
To find the x-intercepts, we set \( f(x) = 0 \), which means we set the numerator equal to zero and solve for \( x \):

\[
2x^2 + 7x + 5 = 0
\]

We solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 7 \), and \( c = 5 \):

\[
x = \frac{-7 \pm \sqrt{7^2 - 4(2)(5)}}{2(2)} = \frac{-7 \pm \sqrt{49 - 40}}{4} = \frac{-7 \pm \sqrt{9}}{4} = \frac{-7 \pm 3}{4}
\]

So the two possible solutions for \( x \) are:

\[
x = \frac{-7 + 3}{4} = \frac{-4}{4} = -1
\]
\[
x = \frac{-7 - 3}{4} = \frac{-10}{4} = -\frac{5}{2}
\]

Thus, the x-intercepts are at \( (-1, 0) \) and \( \left(-\frac{5}{2}, 0\right) \).

### 3) Vertical Asymptotes:
Vertical asymptotes occur where the denominator is zero, but the numerator is not zero at those points. To find the vertical asymptotes, we set the denominator equal to zero and solve for \( x \):

\[
3x^2 - x - 10 = 0
\]

Using the quadratic formula again, with \( a = 3 \), \( b = -1 \), and \( c = -10 \):

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-10)}}{2(3)} = \frac{1 \pm \sqrt{1 + 120}}{6} = \frac{1 \pm \sqrt{121}}{6} = \frac{1 \pm 11}{6}
\]

So the two solutions are:

\[
x = \frac{1 + 11}{6} = \frac{12}{6} = 2
\]
\[
x = \frac{1 - 11}{6} = \frac{-10}{6} = -\frac{5}{3}
\]

Thus, the vertical asymptotes are at \( x = 2 \) and \( x = -\frac{5}{3} \).

### Final Results:
1) The y-intercept is at \( (0, -\frac{1}{2}) \).
2) The x-intercepts are at \( (-1, 0) \) and \( \left(-\frac{5}{2}, 0\right) \).
3) The vertical asymptotes are at \( x = 2 \) and \( x = -\frac{5}{3} \).

Would you like more details or have any questions?

### Related Questions:
1. How do you find horizontal asymptotes for this function?
2. What is the end behavior of \( f(x) \) as \( x \to \infty \)?
3. Can you determine slant asymptotes for \( f(x) \)?
4. How do critical points relate to the graph of this function?
5. How can you graph this rational function based on the intercepts and asymptotes?

### Tip:
Always check both the numerator and the denominator when finding intercepts and asymptotes to ensure that they aren't cancelling each other out.

f(x) = (2x^2+7x+5)/(3x^2-1x-10); This function has: 1) A y-intercept at the point, 2) x-intercepts at the point(s), 3) Vertical asymptotes at x = 

Learn how to find the y-intercept, x-intercepts, and vertical asymptotes for the rational function f(x) = (2x^2 + 7x + 5) / (3x^2 - x - 10). This step-by-step guide uses the quadratic formula and key concepts in rational functions, ideal for Grades 10-12 students.

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