For the funtion x2+x-12/2x2-7x+3          F(5)= ?        vertical asymotome=     Hole=            horizontal/slant asympotome=

Let's break down the function step by step and address each of the following questions: \( F(5) \), vertical asymptotes, holes, and horizontal/slant asymptotes.

Given function: 
\[
F(x) = \frac{x^2 + x - 12}{2x^2 - 7x + 3}
\]

### Step 1: \( F(5) \)
To find \( F(5) \), substitute \( x = 5 \) into the function:

\[
F(5) = \frac{5^2 + 5 - 12}{2(5)^2 - 7(5) + 3}
\]
Let's simplify this:

Numerator: \( 5^2 + 5 - 12 = 25 + 5 - 12 = 18 \)

Denominator: \( 2(5^2) - 7(5) + 3 = 2(25) - 35 + 3 = 50 - 35 + 3 = 18 \)

So,
\[
F(5) = \frac{18}{18} = 1
\]

### Step 2: Vertical Asymptotes
Vertical asymptotes occur where the denominator equals zero, but the numerator does not. We need to find where the denominator \( 2x^2 - 7x + 3 = 0 \).

We solve the quadratic equation:
\[
2x^2 - 7x + 3 = 0
\]
Using the quadratic formula:
\[
x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4}
\]
\[
x = \frac{7 \pm 5}{4}
\]
Thus,
\[
x = \frac{7 + 5}{4} = 3 \quad \text{or} \quad x = \frac{7 - 5}{4} = \frac{1}{2}
\]

So, there are vertical asymptotes at \( x = 3 \) and \( x = \frac{1}{2} \).

### Step 3: Holes
Holes occur where both the numerator and denominator are zero for the same \( x \)-value. To check for holes, we need to factor both the numerator and the denominator.

Factor the numerator:
\[
x^2 + x - 12 = (x - 3)(x + 4)
\]

Factor the denominator:
\[
2x^2 - 7x + 3 = (2x - 3)(x - 1)
\]

Since there are no common factors between the numerator and denominator, there are no holes in this function.

### Step 4: Horizontal/Slant Asymptotes
To find the horizontal or slant asymptote, we compare the degrees of the numerator and the denominator.

- The degree of the numerator is 2 (since \( x^2 \) is the highest term).
- The degree of the denominator is also 2.

When the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 1 (from \( x^2 \)) and the leading coefficient of the denominator is 2 (from \( 2x^2 \)).

Thus, the horizontal asymptote is:
\[
y = \frac{1}{2}
\]

Since there is a horizontal asymptote, there is **no slant asymptote**.

---

### Summary:
- \( F(5) = 1 \)
- Vertical asymptotes: \( x = 3 \) and \( x = \frac{1}{2} \)
- Holes: None
- Horizontal asymptote: \( y = \frac{1}{2} \)
- No slant asymptote.

Would you like further details or have any other questions?

Here are 5 related questions you might consider:
1. How do you find the slant asymptote for a rational function?
2. How can you determine if a function has a removable discontinuity (hole)?
3. What is the process of factoring a quadratic expression like \( 2x^2 - 7x + 3 \)?
4. How does the degree of the numerator and denominator affect asymptotes?
5. Can a function have both a vertical and horizontal asymptote?

**Tip:** When factoring quadratics, always check for common factors in both the numerator and denominator to see if there are any holes.

For the function x^2 + x - 12 / 2x^2 - 7x + 3, F(5) = ? Vertical asymptote = ? Hole = ? Horizontal/Slant asymptote = ?

Solve for F(5) and identify the vertical, horizontal, and slant asymptotes of the rational function x^2 + x - 12 / 2x^2 - 7x + 3. This guide explains key concepts like vertical and horizontal asymptotes, factoring, and the quadratic formula. Ideal for Grades 10-12 students studying algebra.

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