Let \( f(x)=\frac{5 x+2}{2 x^{2}+5 x-25}=\frac{(5 x+2)}{(x+5)(2 x-5)} \) Find: 1) the domain in interval notation Note: Use -oo for \( -\infty \), oo for \( \infty \), U for union. 2) the \( y \) intercept at the point 3) \( x \) intercepts at the point(s) 4) Vertical asymptotes at \( x= \) 5) Horizontal asymptote at \( y=\square \)

To analyze the function \( f(x) = \frac{5x + 2}{2x^2 + 5x - 25} = \frac{5x + 2}{(x + 5)(2x - 5)} \), we will go through each of the requested items step by step. ### 1) Domain in Interval Notation The domain of a rational function is all real numbers except where the denominator is zero. First, we need to find the values of \( x \) that make the denominator zero: \[ 2x^2 + 5x - 25 = 0 \] We can factor this as: \[ (x + 5)(2x - 5) = 0 \] Setting each factor to zero gives us: \[ x + 5 = 0 \quad \Rightarrow \quad x = -5 \] \[ 2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2} \] Thus, the function is undefined at \( x = -5 \) and \( x = \frac{5}{2} \). Therefore, the domain in interval notation is: \[ (-\infty, -5) \cup (-5, \frac{5}{2}) \cup (\frac{5}{2}, \infty) \] ### 2) \( y \) Intercept To find the \( y \)-intercept, we evaluate \( f(0) \): \[ f(0) = \frac{5(0) + 2}{2(0)^2 + 5(0) - 25} = \frac{2}{-25} = -\frac{2}{25} \] Thus, the \( y \)-intercept is at the point \( (0, -\frac{2}{25}) \). ### 3) \( x \) Intercepts To find the \( x \)-intercepts, we set the numerator equal to zero: \[ 5x + 2 = 0 \] Solving for \( x \): \[ 5x = -2 \quad \Rightarrow \quad x = -\frac{2}{5} \] Thus, the \( x \)-intercept is at the point \( \left(-\frac{2}{5}, 0\right) \). ### 4) Vertical Asymptotes Vertical asymptotes occur where the denominator is zero and the numerator is not zero. From our earlier calculations, we found that the vertical asymptotes are at: \[ x = -5 \quad \text{and} \quad x = \frac{5}{2} \] ### 5) Horizontal Asymptote To find the horizontal asymptote, we look at the degrees of the polynomial in the numerator and the denominator. The degree of the numerator (1) is less than the degree of the denominator (2). Therefore, the horizontal asymptote is: \[ y = 0 \] ### Summary of Results 1) Domain: \( (-\infty, -5) \cup (-5, \frac{5}{2}) \cup (\frac{5}{2}, \infty) \) 2) \( y \)-intercept: \( (0, -\frac{2}{25}) \) 3) \( x \)-intercept: \( \left(-\frac{2}{5}, 0\right) \) 4) Vertical asymptotes: \( x = -5 \) and \( x = \frac{5}{2} \) 5) Horizontal asymptote: \( y = 0 \)