x-intercepts (-4,0) and (4, 0) y-intercepts (0,-2) and (0,8)
Practice makes perfect
We are asked to graph the equation x^2+(y-3)^2=25 and state the x- and y-interecepts. Notice that the given equation looks very similar to the standard equation of a circle.
(x- h)^2+(y- k)^2= r^2
Here, the center is the point ( h, k) and the radius is r. To help us graph the given equation, we will rewrite it to match this form, and then we can identify the center and the radius. In this case, we will need to subtract 0 from the x- variable and ensure that the constant term is a square.
As we can see, the equation for the boundary curve matches the standard equation of a circle. The center of this circle is the point ( 0, 3), and its radius is 5. Let's graph it.
Now that we have graphed our equation, we can move to finding the x- and y- intercepts. Looking at the graph we made, it seems that the x-intercepts are the points (-4,0) and (4, 0), while points (-2, 0) and (8,0) are the y-intercepts. Let's plot these points on the graph.
