Question

Identify the x and y intercepts of the graph of the function below. Type your intercepts as a point (x,y). If an intercept does not exist type "none". If more than one intercept exists you can type either intercept.f(x)= -5|x+2|+15 x intercept = Answery intercept = Answer

Answer 1

Let's start with the x -intercepts. The x - intercepts are the points for which y = 0, that is, f(x) = 0:

`\begin{gathered} 0=-5|x+2|+15 \\ 5|x+2|=15 \\ |x+2|=(15)/(5) \\ |x+2|=3 \end{gathered}`

Since the module of x + 2 is 3, the inside can be either equal to 3 or -3, so:

`\begin{gathered} x+2=\pm3 \\ x=\pm3-2 \\ x_1=3-2=1 \\ x_2=-3-2=-5 \end{gathered}`

Thus, the x-intercepts are at x = 1 and x = -5 and, because they intercept x, y = 0, s they are (1, 0) and (-5, 0).

The y - intercept is the value where x = 0, so we can substitute that and sse the value of y, that is, f(0):

`\begin{gathered} f(0)=-5|0+2|+15 \\ f(0)=-5\cdot2+15 \\ f(0)=-10+15 \\ f(0)=5 \end{gathered}`

So, the y-intercept is y = 5, and because it is at x = 0, the point is (0, 5).