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.
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)
• 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)