vertical asymptote at x = - 5 slant asymptote is y = x - 6 Explanation: The denominator of g(x) cannot be zero as this would make g(x) undefined. Equating the denominator to zero and solving gives ...

\displaystyle\lim_{{{x}\to\pm\infty}}{\frac{{{\left({x}-{1}\right)}{\left({x}+{2}\right)}{\left({x}+{5}\right)}}}{{{2}{x}{\left({x}+{2}\right)}}}}=\pm\infty Explanation: First of all, the ...

\displaystyle{x}={8} Explanation: First, we'll cross multiply: if \displaystyle\frac{{a}}{{b}}=\frac{{c}}{{d}} then \displaystyle{a}\cdot{d}={b}\cdot{c} \displaystyle{\left({x}+{12}\right)}{\left({x}-{7}\right)}={\left({x}-{4}\right)}{\left({x}-{3}\right)} ...

Don't Memorise May 9, 2015 \displaystyle{\left(\frac{{1}}{{{2}{x}}}\right)}+{\left(\frac{{3}}{{{x}+{7}}}\right)}=-{\left(\frac{{1}}{{x}}\right)} \displaystyle{\left(\frac{{1}}{{{2}{x}}}\right)}+{\left(\frac{{1}}{{x}}\right)}=-{\left(\frac{{3}}{{{x}+{7}}}\right)} ...

Rachel Aug 18, 2017 \displaystyle{f{{\left({x}\right)}}}=\frac{{{\left({2}{x}-{3}\right)}{\left({x}+{2}\right)}}}{{{x}-{2}}} Asymptotes: "Unreachable value that occurs when a ...

\displaystyle{f{{\left({x}\right)}}}\ \text{ is increasing at }\ {x}=-{1} Explanation: \displaystyle\text{to determine if f(x) is increasing/decreasing at x = a} \displaystyle\text{differentiate and evaluate }\ {f}'{\left({x}\right)}\ \text{ at }\ {x}={a}\text{} ...