1. **State the problem:** Find the intercepts, vertical asymptotes, turning points, and the nature of the curve at the turning points for the function $$y=\frac{4}{(5-x)(x-1)}$$. 2. **Rewrite the function:** The function is $$y=\frac{4}{(5-x)(x-1)} = \frac{4}{-x^2 + 6x - 5}$$. 3. **Find the intercepts:** - **x-intercepts:** Set $$y=0$$, so $$\frac{4}{(5-x)(x-1)}=0$$. Since numerator 4 is never zero, no x-intercepts. - **y-intercept:** Set $$x=0$$, then $$y=\frac{4}{(5-0)(0-1)}=\frac{4}{5 \times (-1)}=\frac{4}{-5}=-\frac{4}{5}$$. 4. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero: $$(5-x)(x-1)=0 \implies x=5 \text{ or } x=1$$. 5. **Find turning points:** Find $$y'$$ and set to zero. Rewrite denominator as $$D = (5-x)(x-1) = -x^2 + 6x -5$$. Then $$y = \frac{4}{D}$$. Derivative: $$y' = -4 \frac{D'}{D^2}$$ Calculate $$D' = \frac{d}{dx}(-x^2 + 6x -5) = -2x + 6$$. Set $$y' = 0$$: $$-4 \frac{-2x + 6}{D^2} = 0 \implies -2x + 6 = 0 \implies x = 3$$. 6. **Find y-coordinate of turning point:** $$y(3) = \frac{4}{(5-3)(3-1)} = \frac{4}{2 \times 2} = 1$$. 7. **Determine nature of turning point:** Use second derivative or test values around $$x=3$$. Second derivative: $$y'' = \frac{d}{dx} y' = \frac{d}{dx} \left(-4 \frac{D'}{D^2}\right)$$ Using quotient and product rules, after simplification: $$y''(3) = \frac{24}{16} = 1.5 > 0$$, so the curve is concave up at $$x=3$$, indicating a local minimum. **Final answers:** - No x-intercepts. - y-intercept at $$\left(0, -\frac{4}{5}\right)$$. - Vertical asymptotes at $$x=1$$ and $$x=5$$. - One turning point at $$\left(3, 1\right)$$ which is a local minimum.