14. Convert the following general form function to vertex form. You must show your work in order to receive credit. f(x)=(8x+30)/(x+3) Did you answer this question on your test paper?

To convert the given function \( f(x) = \frac{8x + 30}{x + 3} \) to vertex form, we need to follow these steps:<br /><br />1. **Rewrite the function in a more convenient form:**<br /> The given function can be rewritten as:<br /> \[<br /> f(x) = \frac{8(x + 3) + 6}{x + 3}<br /> \]<br /> This simplifies to:<br /> \[<br /> f(x) = \frac{8(x + 3)}{x + 3} + \frac{6}{x + 3}<br /> \]<br /> Since \(\frac{8(x + 3)}{x + 3} = 8\), we get:<br /> \[<br /> f(x) = 8 + \frac{6}{x + 3}<br /> \]<br /><br />2. **Express the function in terms of a new variable:**<br /> Let \( u = x + 3 \). Then, \( x = u - 3 \) and the function becomes:<br /> \[<br /> f(u) = 8 + \frac{6}{u}<br /> \]<br /><br />3. **Convert to vertex form:**<br /> The vertex form of a function is typically written as \( y = a(x - h)^2 + k \). However, since our function is not a quadratic but rather a rational function, we can rewrite it in a form that resembles the vertex form by completing the square for the denominator.<br /><br /> We start with:<br /> \[<br /> f(u) = 8 + \frac{6}{u}<br /> \]<br /><br /> To express this in a form similar to \( y = a(x - h)^2 + k \), we recognize that the term \(\frac{6}{u}\) can be rewritten using a hyperbola form:<br /> \[<br /> f(u) = 8 + \frac{6}{u} = 8 + \frac{6}{u}<br /> \]<br /><br /> This does not directly convert to the vertex form \( y = a(x - h)^2 + k \) because it is not a quadratic function. Instead, it is a rational function. Therefore, the original function \( f(x) = \frac{8x + 30}{x + 3} \) does not have a straightforward vertex form representation.<br /><br />In conclusion, the function \( f(x) = \frac{8x + 30}{x + 3} \) cannot be easily converted to the vertex form \( y = a(x - h)^2 + k \) due to its nature as a rational function.