We have the next given function:
[tex]y=x^2+2x+3[/tex]The equation corresponds to a parabola.
First, we need to derivate it:
[tex]\begin{gathered} \frac{d}{dx}y=\frac{d}{dx}(x^2+2x+3) \\ Then \\ \frac{d}{dx}y=\frac{d}{dx}x^2+\frac{d}{dx}2x+\frac{d}{dx}3 \\ \frac{d}{dx}y=2x+2+0 \\ \frac{d}{dx}y=2x+2 \end{gathered}[/tex]Now, we need to find when x=0.
[tex]\begin{gathered} 2x+2=0 \\ Solve\text{ for x} \\ 2x=-2 \\ x=-\frac{2}{2} \\ x=-1 \end{gathered}[/tex]So, we can check the numbers using the number line:
Now, we need to check a number smaller than -1 and another number greater than -1.
Let us check 3.
Then:
[tex]\begin{gathered} 2x+2 \\ 2(3)+2 \\ 6+2 \\ 8 \end{gathered}[/tex]POSITIVE, when the x value is greater than -1, the function is positive (it is increasing).
Let us check -2:
[tex]\begin{gathered} 2x+2 \\ 2(-2)+2 \\ -4+2 \\ -2 \end{gathered}[/tex]NEGATIVE, when the x value is smaller than -1, the function is decreasing.
Finally, we can write each interval.
The function is decreasing on interval (-∞,-1)
The function is increasing on interval (-1,∞)
