3) Write an equation for a rational function whose graph has all of the indicated features.

• X-intercept of 5/3.
• Y-intercept of – ½.
• Vertical asymptote with equation x = -10.
• Horizontal asymptote with equation y = 3.

2 answers

• X-intercept of 5/3
y = (3x-5)

• Vertical asymptote with equation x = -10
y = (3x-5)/(x+10)

• Horizontal asymptote with equation y = 3
y = 3(3x-5)/(x+10)

• Y-intercept of -1/2
Now we have a problem. Our y-intercept is at y = -3/2.
We can't just shift it up, since that will change the x-intercept.
We can't just multiply by -1/6 because that will change the horizontal asymptote.
So, how can we change the value only at x=0? Consider this:
y = 3(3x-5)(x^2+1) / (x+10)(x^2+3)
The two new factors don't add any zeroes or asymptotes, because they are never zero. But they do change the value of y(0)
Did you catch my mistake?
Here's solution 2.0

• X-intercept of 5/3
y = (3x-5)

• Vertical asymptote with equation x = -10
y = (3x-5)/(x+10)

• Horizontal asymptote with equation y = 3
✅ equal degree, ratio of coefficients is 3

• Y-intercept of -1/2
✅ y(0) = -1/2

Looks like you have a solution:
y = (3x-5)/(x+10)