graph the function y=x^4 - 27x^2 =x^2 (x^2 - 27) by identifying the domain and any symmetries, finding the…

Explanation:

Step1: Find the domain

The function $y = x^{4}-27x^{2}$ is a polynomial. So the domain is $(-\infty,\infty)$.

Step2: Check for symmetries

Replace $x$ with $-x$: $y=(-x)^{4}-27(-x)^{2}=x^{4}-27x^{2}$. The function is even, so it is symmetric about the y - axis.

Step3: Find the first - derivative

Using the power rule, if $y = x^{4}-27x^{2}$, then $y'=4x^{3}-54x = 2x(2x^{2}-27)$.

Step4: Find the critical points

Set $y' = 0$. Then $2x(2x^{2}-27)=0$. $2x=0$ gives $x = 0$, and $2x^{2}-27=0$ gives $x=\pm\frac{3\sqrt{6}}{2}$.

Step5: Analyze increasing and decreasing intervals

Test intervals: For $x<-\frac{3\sqrt{6}}{2}$, let $x=-3\sqrt{2}$, then $y'=2(-3\sqrt{2})[2(-3\sqrt{2})^{2}-27]=2(-3\sqrt{2})(36 - 27)<0$, so the function is decreasing on $(-\infty,-\frac{3\sqrt{6}}{2})$. For $-\frac{3\sqrt{6}}{2}<x<0$, let $x = - 1$, then $y'=2(-1)[2(-1)^{2}-27]=2(-1)(2 - 27)>0$, so the function is increasing on $(-\frac{3\sqrt{6}}{2},0)$. For $0<x<\frac{3\sqrt{6}}{2}$, let $x = 1$, then $y'=2(1)[2(1)^{2}-27]=2(2 - 27)<0$, so the function is decreasing on $(0,\frac{3\sqrt{6}}{2})$. For $x>\frac{3\sqrt{6}}{2}$, let $x = 3\sqrt{2}$, then $y'=2(3\sqrt{2})[2(3\sqrt{2})^{2}-27]=2(3\sqrt{2})(36 - 27)>0$, so the function is increasing on $(\frac{3\sqrt{6}}{2},\infty)$.

Step6: Find the second - derivative

$y'=4x^{3}-54x$, then $y'' = 12x^{2}-54=6(2x^{2}-9)$.

Step7: Find the inflection points

Set $y'' = 0$. Then $6(2x^{2}-9)=0$, which gives $x=\pm\frac{3}{\sqrt{2}}$.

Step8: Analyze concavity

For $x<-\frac{3}{\sqrt{2}}$, let $x=-2$, then $y''=12(-2)^{2}-54 = 48 - 54<0$, so the function is concave down on $(-\infty,-\frac{3}{\sqrt{2}})$. For $-\frac{3}{\sqrt{2}}<x<\frac{3}{\sqrt{2}}$, let $x = 0$, then $y''=12(0)^{2}-54=-54<0$, so the function is concave down on $(-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}})$. For $x>\frac{3}{\sqrt{2}}$, let $x = 2$, then $y''=12(2)^{2}-54 = 48 - 54>0$, so the function is concave up on $(\frac{3}{\sqrt{2}},\infty)$.

Step9: Find the intercepts

When $x = 0$, $y = 0$. When $y = 0$, $x^{2}(x^{2}-27)=0$, so $x = 0,\pm3\sqrt{3}$.

Step10: Find the absolute extreme points

Since $\lim_{x\rightarrow\pm\infty}y=\lim_{x\rightarrow\pm\infty}(x^{4}-27x^{2})=\infty$. $y(-\frac{3\sqrt{6}}{2})=y(\frac{3\sqrt{6}}{2})=(\frac{3\sqrt{6}}{2})^{4}-27(\frac{3\sqrt{6}}{2})^{2}=\frac{324}{4}\times\frac{9}{4}-27\times\frac{54}{4}=\frac{2916 - 5832}{16}=-\frac{2916}{16}=-\frac{729}{4}$ and $y(0)=0$. The absolute minimum points are $(-\frac{3\sqrt{6}}{2},-\frac{729}{4})$ and $(\frac{3\sqrt{6}}{2},-\frac{729}{4})$.

Answer:

Domain: $(-\infty,\infty)$; Symmetry: Symmetric about the y - axis; Critical points: $x = 0,\pm\frac{3\sqrt{6}}{2}$; Increasing intervals: $(-\frac{3\sqrt{6}}{2},0),(\frac{3\sqrt{6}}{2},\infty)$; Decreasing intervals: $(-\infty,-\frac{3\sqrt{6}}{2}),(0,\frac{3\sqrt{6}}{2})$; Inflection points: $x=\pm\frac{3}{\sqrt{2}}$; Concave down intervals: $(-\infty,-\frac{3}{\sqrt{2}}),(-\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}})$; Concave up intervals: $(\frac{3}{\sqrt{2}},\infty)$; x - intercepts: $x = 0,\pm3\sqrt{3}$; y - intercept: $y = 0$; Absolute minimum points: $(-\frac{3\sqrt{6}}{2},-\frac{729}{4})$ and $(\frac{3\sqrt{6}}{2},-\frac{729}{4})$