如何构造具有指定垂直渐近线、斜渐近线与y截距的有理函数g(x)？
Let's walk through building this rational function ( g(x) ) step by step—each requirement gives us a clear hint about its structure:
Vertical asymptotes at ( x=2 ) and ( x=5 ) mean the denominator must have factors that equal zero exactly at these points (and the numerator can't be zero there). So we start with:Denominator = (x-2)(x-5) = x^2 - 7x + 10
A slant asymptote ( y=2x+1 ) tells us two key things about our rational function:
- The degree of the numerator is exactly 1 higher than the denominator (denominator is degree 2, so numerator is degree 3).
- As ( x \to \infty ), ( g(x) ) approaches ( 2x+1 ).
We can express ( g(x) ) as the sum of the slant asymptote and a "remainder" term that vanishes at infinity:g(x) = (2x + 1) + \frac{C}{(x-2)(x-5)}
Here, ( C ) is a constant (we don't need a higher-degree remainder because any lower-degree polynomial than the denominator will tend to 0 as ( x ) gets large—keeping it a constant simplifies calculations).
The y-intercept is 7, which means ( g(0) = 7 ). Plug ( x=0 ) into our expression:
g(0) = (2*0 + 1) + \frac{C}{(0-2)(0-5)} = 1 + \frac{C}{10} = 7
Solve for ( C ):\frac{C}{10} = 7 - 1 = 6 → C = 60
Substitute ( C=60 ) back into our expression. You can leave it in split form for clarity:g(x) = 2x + 1 + \frac{60}{(x-2)(x-5)}
Or combine it into a single rational expression by expanding the numerator:
g(x) = \frac{(2x+1)(x-2)(x-5) + 60}{(x-2)(x-5)}
Expanding the numerator gives:2x^3 - 13x^2 + 13x + 70
So the single-fraction form is:g(x) = \frac{2x^3 - 13x^2 + 13x + 70}{x^2 - 7x + 10}
Let's double-check all requirements are met:
- Vertical Asymptotes: Denominator is zero at ( x=2 ) and ( x=5 ); numerator at these points is 60 (non-zero), so asymptotes are valid.
- Slant Asymptote: As ( x \to \infty ), the fraction ( \frac{60}{(x-2)(x-5)} ) approaches 0, so ( g(x) \to 2x+1 ).
- y-Intercept: Plugging ( x=0 ) into the single fraction gives ( \frac{70}{10} =7 ), which matches the requirement.
内容的提问来源于stack exchange，提问作者CodeMan5000
