Write an equation for a rational function with:

- Vertical asymptotes at [tex]\(x = -5\)[/tex] and [tex]\(x = 6\)[/tex]
- [tex]\(x\)[/tex]-intercepts at [tex]\(x = 1\)[/tex] and [tex]\(x = 4\)[/tex]
- [tex]\(y\)[/tex]-intercept at 5

[tex]\[y = \, \square \][/tex]



Answer :

Certainly! Let's write an equation for a rational function that satisfies the following conditions:
1. Vertical asymptotes at [tex]\( x = -5 \)[/tex] and [tex]\( x = 6 \)[/tex]
2. [tex]\( x \)[/tex]-intercepts at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex]
3. [tex]\( y \)[/tex]-intercept at 5

### Step-by-Step Solution:

1. Vertical Asymptotes:
The vertical asymptotes occur where the denominator of the rational function is zero. Therefore, the denominator must have factors that become zero at [tex]\( x = -5 \)[/tex] and [tex]\( x = 6 \)[/tex]. Hence, the denominator can be written as:
[tex]\[ D(x) = (x + 5)(x - 6) \][/tex]

2. [tex]\( x \)[/tex]-Intercepts:
The [tex]\( x \)[/tex]-intercepts occur where the numerator of the rational function is zero. Therefore, the numerator must have factors that become zero at [tex]\( x = 1 \)[/tex] and [tex]\( x = 4 \)[/tex]. Hence, the numerator can be written as:
[tex]\[ N(x) = (x - 1)(x - 4) \][/tex]

3. Forming the Rational Function:
Combining the numerator and the denominator, we have:
[tex]\[ R(x) = \frac{N(x)}{D(x)} = \frac{(x - 1)(x - 4)}{(x + 5)(x - 6)} \][/tex]
At this stage, the rational function does not yet account for the [tex]\( y \)[/tex]-intercept.

4. Adjusting for the [tex]\( y \)[/tex]-Intercept:
To ensure the [tex]\( y \)[/tex]-intercept is 5 (i.e., [tex]\( R(0) = 5 \)[/tex]), we need to determine a constant factor [tex]\( k \)[/tex] that will adjust the function accordingly.

First, find the value of the function at [tex]\( x = 0 \)[/tex] without the constant:
[tex]\[ R(0) = \frac{(0 - 1)(0 - 4)}{(0 + 5)(0 - 6)} = \frac{(-1)(-4)}{(5)(-6)} = \frac{4}{-30} = -\frac{2}{15} \][/tex]

To achieve the [tex]\( y \)[/tex]-intercept of 5, we need:
[tex]\[ k \cdot \left(-\frac{2}{15}\right) = 5 \][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[ k = 5 \cdot \left(-\frac{15}{2}\right) = -\frac{75}{2} \][/tex]

5. Final Rational Function:
The final rational function, taking into account the constant [tex]\( k \)[/tex], is:
[tex]\[ R(x) = k \cdot \frac{(x - 1)(x - 4)}{(x + 5)(x - 6)} = -\frac{75}{2} \cdot \frac{(x - 1)(x - 4)}{(x + 5)(x - 6)} \][/tex]

Simplifying, we get:
[tex]\[ R(x) = \frac{-75 (x - 1) (x - 4)}{2 (x + 5) (x - 6)} \][/tex]

### Final Answer:
[tex]\[ y = \frac{-75 (x - 1) (x - 4)}{2 (x + 5) (x - 6)} \][/tex]