Question

построить график функции y=y = |x^2+4|x|-5|

Answer 1

$\displaystyle\left[\begin{gathered}\begin{cases}x\geq0\\y=|x^2+4x-5|\end{cases}\\\begin{cases}x<0\\y=|x^2-4x-5|\end{cases}\end{gathered}\to\left[\begin{gathered}\begin{cases}x\geq0\\y=|x^2+4x-5|\end{cases}\\\begin{cases}x<0\\y=|x^2-4x-5|\end{cases}\end{gathered}$

$x^2+4x-5=0\\x_1=-5;x_2=1\\\\\\x^2-4x-5\\x_1=-1;x_2=5$

$\displaystyle\left[\begin{gathered}\begin{cases}x\geq0\\x<-5;x\geq1\\y=x^2+4x-5\end{cases}\\\begin{cases}x\geq0\\-5\leq x<1\\y=-x^2-4x+5\end{cases}\end{gathered}\to\left[\begin{gathered}\begin{cases}x\geq1\\y=x^2+4x-5\end{cases}\\\begin{cases}0\leq x<1\\y=-x^2-4x+5\end{cases}\end{gathered}$

$\displaystyle\left[\begin{gathered}\begin{cases}x<0\\x<-1;x\geq5\\y=x^2-4x-5\end{cases}\\\begin{cases}x<0\\-1\leq x<5\\y=-x^2+4x+5\end{cases}\end{gathered}\to\left[\begin{gathered}\begin{cases}x<-1\\y=x^2-4x-5\end{cases}\\\begin{cases}-1\leq x<0\\y=-x^2+4x+5\end{cases}\end{gathered}$

Согласно данных двух совокупностей строим график (вложение)