Let \( f(x)=\frac{3 x^{2}-5 x-12}{2 x^{2}-9 x-5} \) This function has: 1) A \( y \) intercept at the point 2) \( x \) intercepts at the point(s) \( \square \) 3) Vertical asymptotes at \( x= \) \( \square \) 4) Horizontal asymptote at \( y= \) \( \square \) Question Help: Video Calculator

To find the \( y \)-intercept, set \( x = 0 \) in the function \( f(x) \). This gives \( f(0) = \frac{3(0)^2 - 5(0) - 12}{2(0)^2 - 9(0) - 5} = \frac{-12}{-5} = \frac{12}{5} \). Thus, the \( y \)-intercept is at the point \( (0, \frac{12}{5}) \). Next, to find the \( x \)-intercepts, set \( f(x) = 0 \), which means setting the numerator equal to zero: \( 3x^2 - 5x - 12 = 0 \). Factoring (or using the quadratic formula), we find the roots are at \( x = 4 \) and \( x = -1 \). Therefore, the \( x \)-intercepts are at the points \( (4, 0) \) and \( (-1, 0) \). For vertical asymptotes, we look at where the denominator is zero: \( 2x^2 - 9x - 5 = 0 \). Factoring (or using the quadratic formula) gives the solutions \( x = 5 \) and \( x = -\frac{1}{2} \), so the vertical asymptotes are at \( x = 5 \) and \( x = -\frac{1}{2} \). Lastly, to determine the horizontal asymptote, we compare the leading coefficients of the numerator and denominator. Since both the numerator and denominator are quadratic, the horizontal asymptote is given by the ratio of the leading coefficients: \( y = \frac{3}{2} \). So, summarizing: 1) The \( y \) intercept is at \( (0, \frac{12}{5}) \). 2) The \( x \) intercepts are at the points \( (4, 0) \) and \( (-1, 0) \). 3) The vertical asymptotes are at \( x = 5 \) and \( x = -\frac{1}{2} \). 4) The horizontal asymptote is at \( y = \frac{3}{2} \).