sketch-the-curve-of-y-x-2-1-x-2-7x-12-

Question Number 123199 by aurpeyz last updated on 23/Nov/20

s k e t c h t h e c u r v e o f y = \frac{x^{2} - 1}{x^{2} - 7 x - 12}

Answered by MJS_new last updated on 24/Nov/20

y = \frac{x^{2} - 1}{x^{2} + 7 x - 12}

y^{'} = - \frac{7 x^{2} + 22 x + 7}{{(x^{2} + 7 x - 12)}^{2}}

y^{″} = \frac{2 (7 x^{3} + 33 x^{2} + 21 x + 83)}{{(x^{2} - 7 x - 12)}^{3}}

y = 0 \Rightarrow zeros at x = \pm 1

x^{2} + 7 x - 12 = 0 \Rightarrow asymptotes at x = \frac{7 \pm \sqrt{97}}{2} (\approx - 1.42 and 8.42)

y = 1 + \frac{7 x + 11}{x^{2} - 7 x - 12} \Rightarrow asymptote at y = 1 (for x = \pm \infty)

y^{'} = 0 \Rightarrow extremes at x = - \frac{11 \pm 6 \sqrt{2}}{7} (\approx - 2.78 and - . 36)

y^{″} {\begin{cases} > 0 for x = - \frac{11 + 6 \sqrt{2}}{7} \Rightarrow local \min at \approx (\begin{array}{c} - 2.78 \\ . 44 \end{array}) \\ < 0 for x = - \frac{11 - 6 \sqrt{2}}{7} \Rightarrow local \max at \approx (\begin{array}{c} - . 36 \\ . 09 \end{array}) \end{cases}

y^{″} = 0 \Rightarrow inflection point \approx (\begin{matrix} - 4.62 \\ . 49 \end{matrix})

y^{″} {\begin{cases} < 0 for x <\approx - 4.62 \Rightarrow curvature - \\ > 0 for \approx - 4.62 < x <\approx - 1.42 \Rightarrow curvature + \\ < 0 for \approx - 1.42 < x <\approx 8.42 \Rightarrow curvature - \\ > 0 for x >\approx 8.42 \Rightarrow curvature + \end{cases}

now sketch it

Facebook Tweet Pin