How to Calculate and Interpret Intersection Points

Since $f(x)=y$ and $g(x)=y$, you can set them equal to each other:

$$\begin{array}{llll}\hfill -x-1& =2x+8& \hfill & \\ \hfill -3x& =9|÷-3& \hfill & \\ \hfill x& =-3& \hfill & \end{array}$$

Insert $x$ into $f(x)=-x-1$ because it’s the simpler expression of the two. You could also insert $x$ into $g(x)$, but it would be more work.

$$f(-3)=-(-3)-1=2.$$

The intersection between $f(x)$ and $g(x)$ is

$$(x,y)=(-3,2).$$

Since $f(x)=y$ and $g(x)=y$, you can set them equal to each other:

$$\begin{array}{llll}\hfill x^2+3x-2& =2x+3& \hfill & \\ \hfill x^2+x-5& =0& \hfill & \end{array}$$

Solving the quadratic equation with the quadratic formula:

$$\begin{array}{llll}\hfill x& =\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot (-5)}}{2\cdot 1}& \hfill & \\ \hfill & =\frac{-1\pm \sqrt{1+20}}{2}& \hfill & \\ \hfill & =\frac{-1\pm \sqrt{21}}{2},& \hfill & \end{array}$$

so

$$\begin{array}{llll}\hfill x_1& =\frac{-1-\sqrt{21}}{2}\approx -2.79,& \hfill & \\ \hfill x_2& =\frac{-1+\sqrt{21}}{2}\approx 1.79.& \hfill & \end{array}$$

Insert $x_1$ and $x_2$ into $g(x)=2x+3$ because it’s the simpler expression. You could also insert $x_1$ and $x_2$ into $f(x)$, but it would be more work.

The intersection between $f(x)$ and $g(x)$ is

$$\begin{array}{llll}\hfill (x_1,y_1)& =(-2.79,-2.58),& \hfill & \\ \hfill (x_2,y_2)& =(1.79,6.58).& \hfill & \end{array}$$

Find the point of intersection between the graphs by setting them equal to each other:

$$\begin{array}{llll}\hfill \sin x& =\cos x|÷\cos x& \hfill & \\ \hfill \tan x& =1& \hfill & \\ \hfill x& ={\tan }^{-1}(1)=\frac{\pi }{4}+n\pi & \hfill & \end{array}$$

You need to check that you didn’t miss any solutions since you divided by $\cos x$, which can be 0. You check this by looking at what happens if $\cos x=0$. Then $x=\frac{\pi }{2}$, which means $\sin x=1$, or $x=\frac{3\pi }{2}$, which means $\sin x=-1$. In both cases, $\sin x$ is different from $\cos x$, so they are not solutions.

There are infinitely many points of intersection, and the points have $x$-values equal to $x=\frac{\pi }{4}+n\pi$ for $n\in \mathbb{Z}$. You find the $y$-values by inserting the $x$-values into one of the functions, for example to $f(x)=\sin x$. Here, you have to be aware that even if $\tan x$ has a period of $\pi$, you have found two different angles on the unit circle, with $\pi$ radians between. This is why you get two different values when you put the result back into $f(x)=\sin x$:

$$\begin{array}{llll}\hfill f\left(\frac{\pi }{4}\right)& =\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}& \hfill & \\ \hfill f\left(\frac{\pi }{4}+\pi \right)& =\sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}& \hfill & \end{array}$$

These are two different $y$-values, $y=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$, each belonging to their respective angle on the unit circle. These angles repeat themselves with a period of $2\pi$. Together, you get two different points of intersection:

Note! In this type of exercise, you are pleased with an answer containing $n$. This is because your were asked to find all the points, meaning the general solution, not the points within a certain interval.