Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $8$
The four x-intercepts are $\pm \sqrt{4 - 2\sqrt{2}}$ and $\pm \sqrt{4 + 2\sqrt{2}}$
C. The function is an even function.
D. $\lim\limits_{x \to -\infty} (x^4-8x^2+8) = \infty$
$\lim\limits_{x \to \infty} (x^4-8x^2+8) = \infty$
There are no asymptotes.
E. The function is increasing on the intervals $(-2,0)\cup (2, \infty)$
The function is decreasing on the intervals $(-\infty,-2)\cup (0,2)$
F. $(-2,-8)$ and $(2, -8)$ are local minima.
$(0, 8)$ is the local maximum.
G. The graph is concave down on the interval $(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3})$
The graph is concave up on the intervals $(-\infty, -\frac{2\sqrt{3}}{3}) \cup (\frac{2\sqrt{3}}{3}, \infty)$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{8}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{8}{9})$
H. We can see a sketch of the curve below.
Work Step by Step
$y = x^4-8x^2+8$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = (0)^4-8(0)^2+8 = 8$
The y-intercept is $8$
When $y = 0$:
$x^4-8x^2+8 = 0$
We can use the quadratic formula:
$x^2 = \frac{8\pm \sqrt{(-8)^2-(4)(1)(8)}}{2(1)}$
$x^2 = \frac{8\pm \sqrt{32}}{2}$
$x^2 = 4 \pm 2\sqrt{2}$
$x = \pm \sqrt{4 \pm 2\sqrt{2}}$
The four x-intercepts are $\pm \sqrt{4 - 2\sqrt{2}}$ and $\pm \sqrt{4 + 2\sqrt{2}}$
C. $(-x)^4-8(-x)^2+8 = x^4-8x^2+8$
The function is an even function.
D. $\lim\limits_{x \to -\infty} (x^4-8x^2+8) = \infty$
$\lim\limits_{x \to \infty} (x^4-8x^2+8) = \infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = 4x^3-16x = 0$
$4x(x^2-4) = 0$
$x = 0, -2, 2$
The function is increasing on the intervals $(-2,0)\cup (2, \infty)$
The function is decreasing on the intervals $(-\infty,-2)\cup (0,2)$
F. When $x = -2$, then $y = (-2)^4-8(-2)^2+8= -8$
When $x = 2$, then $y = (2)^4-8(2)^2+8= -8$
$(-2,-8)$ and $(2, -8)$ are local minima.
When $x = 0$, then $y = (0)^4-8(0)^2+8= 8$
$(0, 8)$ is the local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = 12x^2-16 = 0$
$3x^2-4 = 0$
$x = \pm \frac{2\sqrt{3}}{3}$
When $-\frac{2\sqrt{3}}{3} \lt x \lt \frac{2\sqrt{3}}{3}$, then $y'' \lt 0$
The graph is concave down on the interval $(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3})$
When $x \lt -\frac{2\sqrt{3}}{3}$ or $x \gt \frac{2\sqrt{3}}{3}$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty, -\frac{2\sqrt{3}}{3}) \cup (\frac{2\sqrt{3}}{3}, \infty)$
When $x = -\frac{2\sqrt{3}}{3}$, then $y = (-\frac{2\sqrt{3}}{3})^4-8(-\frac{2\sqrt{3}}{3})^2+8= -\frac{8}{9}$
When $x = \frac{2\sqrt{3}}{3}$, then $y = (\frac{2\sqrt{3}}{3})^4-8(\frac{2\sqrt{3}}{3})^2+8= -\frac{8}{9}$
The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{8}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{8}{9})$
H. We can see a sketch of the curve below.
