Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $5$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (-2x^3-3x^2+12x+5) = \infty$
$\lim\limits_{x \to \infty} (-2x^3-3x^2+12x+5) = -\infty$
There are no asymptotes.
E. The function is decreasing on the intervals $(-\infty,-2)\cup (1,\infty)$
The function is increasing on the interval $(-2,1)$
F. The local minimum is $(-2,-15)$
The local maximum is $(1,12)$
G. The graph is concave down on the interval $(-\frac{1}{2}, \infty)$
The graph is concave up on the interval $(-\infty, -\frac{1}{2})$
The point of inflection is $(-0.5, -1.5)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = -2x^3-3x^2+12x+5$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = -2(0)^3-3(0)^2+12(0)+5 = 5$
The y-intercept is $5$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (-2x^3-3x^2+12x+5) = \infty$
$\lim\limits_{x \to \infty} (-2x^3-3x^2+12x+5) = -\infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = -6x^2-6x+12 = 0$
$x^2+x-2 = 0$
$(x+2)(x-1) = 0$
$x = -2, 1$
When $x\lt -2$ or $x \gt 1$, then $y' \lt 0$
The function is decreasing on the intervals $(-\infty,-2)\cup (1,\infty)$
When $-2 \lt x \lt 1$, then $y' \gt 0$
The function is increasing on the interval $(-2,1)$
F. When $x = -2,$ then $y = -2(-2)^3-3(-2)^2+12(-2)+5 = -15$
The local minimum is $(-2,-15)$
When $x = 1,$ then $y = -2(1)^3-3(1)^2+12(1)+5 = 12$
The local maximum is $(1,12)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -12x-6 = 0$
$2x+1 = 0$
$x = -\frac{1}{2}$
When $x \gt -\frac{1}{2}$, then $y'' \lt 0$
The graph is concave down on the interval $(-\frac{1}{2}, \infty)$
When $x \lt -\frac{1}{2}$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, -\frac{1}{2})$
When $x=-\frac{1}{2}$, then $y = -2(-\frac{1}{2})^3-3(-\frac{1}{2})^2+12(-\frac{1}{2})+5 = -1.5$
The point of inflection is $(-0.5, -1.5)$
H. We can see a sketch of the curve below.
