Find all asymptotes, \( x \)-intercepts, and \( y \)-intercepts (if they do not exist, answer "DNE") for the graph of the rational function and sketch the graph of the function. \( f(x)=\frac{3 x}{x^{2}-8 x+16} \) \( x=\square \) \( y= \) \( x \)-intercept(s): \( y \)-intercept(s):

To analyze the rational function \( f(x) = \frac{3x}{x^2 - 8x + 16} \), we will find the asymptotes, \( x \)-intercepts, and \( y \)-intercepts. ### Step 1: Find the \( x \)-intercepts The \( x \)-intercepts occur when \( f(x) = 0 \). This happens when the numerator is zero (as long as the denominator is not zero at the same point). Set the numerator equal to zero: \[ 3x = 0 \implies x = 0 \] Thus, the \( x \)-intercept is: \[ x\text{-intercept: } (0, 0) \] ### Step 2: Find the \( y \)-intercept The \( y \)-intercept occurs when \( x = 0 \): \[ f(0) = \frac{3(0)}{0^2 - 8(0) + 16} = \frac{0}{16} = 0 \] Thus, the \( y \)-intercept is: \[ y\text{-intercept: } (0, 0) \] ### Step 3: Find the vertical asymptotes Vertical asymptotes occur where the denominator is zero (and the numerator is not zero at those points). We need to solve: \[ x^2 - 8x + 16 = 0 \] Factoring the quadratic: \[ (x - 4)^2 = 0 \implies x = 4 \] Thus, there is a vertical asymptote at: \[ x = 4 \] ### Step 4: Find the horizontal asymptote To find the horizontal asymptote, we look at the degrees of the numerator and denominator. The degree of the numerator (which is 1) is less than the degree of the denominator (which is 2). Therefore, the horizontal asymptote is: \[ y = 0 \] ### Summary of Results - \( x \)-intercept(s): \( (0, 0) \) - \( y \)-intercept(s): \( (0, 0) \) - Vertical asymptote: \( x = 4 \) - Horizontal asymptote: \( y = 0 \) ### Sketching the Graph 1. The graph passes through the origin (0, 0). 2. There is a vertical asymptote at \( x = 4 \), meaning the graph will approach this line but never touch it. 3. The graph approaches the horizontal asymptote \( y = 0 \) as \( x \) goes to \( \pm \infty \). The graph will look like this: - For \( x < 4 \), the function is positive and approaches \( y = 0 \) as \( x \) goes to \( -\infty \). - For \( x > 4 \), the function is negative and approaches \( y = 0 \) as \( x \) goes to \( +\infty \). ### Final Answer - \( x \)-intercept(s): \( (0, 0) \) - \( y \)-intercept(s): \( (0, 0) \) - Vertical asymptote: \( x = 4 \) - Horizontal asymptote: \( y = 0 \)