# Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 321: 1

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $0$ The x-intercepts are $0$ and $-3$ C. The function is not odd or even. D. There are no asymptotes. E. The function is increasing on the intervals $(-\infty, -2)\cup (0, \infty)$ The function is decreasing on the interval $(-2,0)$ F. $(0,0)$ is a local minimum. $(-2,4)$ is a local maximum. G. The graph is concave down on the interval $(-\infty, -1)$ The graph is concave up on the interval $(-1,\infty)$ The point of inflection is $(-1, 2)$ H. We can see a sketch of the curve below. $y = x^3+3x^2$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x = 0,$ then $y = (0)^3+3(0)^2$ = 0 The y-intercept is $0$ When $y = 0$: $x^3+3x^2 = 0$ $x^2(x+3) = 0$ $x = 0, -3$ The x-intercepts are $0$ and $-3$ C. The function is not odd or even. D. $\lim\limits_{x \to -\infty} (x^3+3x^2) = \lim\limits_{x \to -\infty} (x^2)(x+3) = -\infty$ $\lim\limits_{x \to \infty} (x^3+3x^2) = \infty$ There are no asymptotes. E. We can find the values of $x$ such that $y' = 0$: $y' = 3x^2+6x = 0$ $3x(x+2) = 0$ $x = 0, -2$ The function is increasing on the intervals $(-\infty, -2)\cup (0, \infty)$ The function is decreasing on the interval $(-2,0)$ F. When $x = 0$, then $y = (0)^3+3(0)^2 = 0$ $(0,0)$ is a local minimum. When $x = -2$, then $y = (-2)^3+3(-2)^2 = 4$ $(-2,4)$ is a local maximum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = 6x+6 = 0$ $x+1 = 0$ $x = -1$ When $x \lt -1$, then $y'' \lt 0$ The graph is concave down on the interval $(-\infty, -1)$ When $x \gt -1$, then $y'' \gt 0$ The graph is concave up on the interval $(-1,\infty)$ When $x=-1$, then $y = (-1)^3+3(-1)^2 = 2$ The point of inflection is $(-1, 2)$ H. We can see a sketch of the curve below. 