Find the slant asymptote and the vertical asymptote. (If an answer does not exist, enter DNE, ) \[ (x)=\frac{x^{2}}{x-7} \] slant asymptote vertical asymptote

**Step 1: Identify the vertical asymptote** The function is given by \[ f(x)=\frac{x^2}{x-7}. \] A vertical asymptote occurs when the denominator is zero (and the numerator is nonzero). Set the denominator equal to zero: \[ x-7=0. \] Solving for \(x\), we get: \[ x=7. \] Thus, the vertical asymptote is at \(x=7\). **Step 2: Find the slant asymptote** Since the degree of the numerator (which is \(2\)) is exactly one more than the degree of the denominator (which is \(1\)), there is a slant asymptote. To find it, perform polynomial long division by dividing \(x^2\) by \(x-7\). 1. Divide \(x^2\) by \(x\) to get \(x\). 2. Multiply \(x\) by \(x-7\) to get: \[ x(x-7)=x^2-7x. \] 3. Subtract the result from \(x^2\): \[ x^2 - (x^2-7x)=7x. \] 4. Divide \(7x\) by \(x\) to get \(7\). 5. Multiply \(7\) by \(x-7\) to get: \[ 7(x-7)=7x-49. \] 6. Subtract: \[ 7x - (7x-49)=49. \] Thus, the division yields: \[ \frac{x^2}{x-7} = x+7+\frac{49}{x-7}. \] As \(x\) approaches \(\pm\infty\), the term \(\frac{49}{x-7}\) tends to \(0\). Therefore, the slant asymptote is: \[ y=x+7. \] **Final Answers:** - **Slant asymptote:** \(y=x+7\) - **Vertical asymptote:** \(x=7\)