# $g(x) = 5(x+3)^2-2$ determine the interval(s) on which the function is increasing and decreasing

I'm working through a textbook exercise and am asked to find the intervals in which $$g(x) = 5(x+3)^2-2$$ is increasing and decreasing.

The solution provided is decreasing on $$(-\infty, -3)$$ and then increasing on $$(-3, \infty)$$.

I cannot see how to arrive at this conclusion. These are exercises at the end of a chapter on the transformation of functions though the author may be including bits from previous chapters, however I searched through recent chapters and there was only a section on determining where a function is increasing or decreasing based on a graph, so visually.

In this case does the question expect me to derive the solution based on formula? Is that possible? Or am I perhaps expected to draw a graph for some arbitrary values and then try to determine the increasing and decreasing points?

• Did you mean $5(x\color{red}-3)^2-2,$ or did you mean $(-\infty,\color{red}-3)$ and $(\color{red}-3,\infty)$? – J. W. Tanner Sep 24 at 14:24
• Hi thanks for pointing that out, I have updated my question to match the textbook now. – Doug Fir Sep 24 at 14:32

The $$-2$$ at the end is just translating the function down, it will not change if it is increasing or decreasing. Than you have a parabola, so you need to look at the vertex, which is at $$-3$$, since you have $$(x-(-3))^2$$. This is similar to $$x^2$$, just translated horizontally from $$0$$ to $$-3$$. You know that the parabola has values of $$+\infty$$ at $$x=\pm\infty$$, so it's decreasing from $$-\infty$$ to the vertex at $$-3$$, and increasing afterwards.
Hint: $$g'(x)=10(x+3)$$ so you have to solve $$g'(x)\geq 0$$ and $$g'(x)<0$$So compute $$g(x_2)-g(x_1)$$ with h $$x_1
If you know (or see) that translations do not modify monotonicity, you could start from the function $$Y=5X^2$$ and put $$X=x-3$$ and $$Y=y+2$$. A translation in the $$y$$ variable does not alter monotonicity inteervals. A translation in the $$x$$ variable just translates accordingly the monotonicity intervals. Since $$Y=5X^2$$ is decreasin on $$(-\infty,0)$$ and increasing on $$(0,+\infty)$$, then $$y=5(x-3)^2-2$$ is decreasing on ... and increasing on ...