Question

find the points of intersection of the graph of y=-lxl and he circle x^2+y^2=.

Answer 1

what is the radius of the circle

Answer 2

the graph y=-|x| where x will always be positive within the absolute value so you can write it such as y=-(x) = -x

Answer 3

however i don't have the full circle equation so i can't tell you where they intersect

Answer 4

\[x^2+y^2=r^2\]

\[y=\sqrt{r^2-x^2}\]

\[\sqrt{r^2-x^2}=-x\]

Answer 5

\[r^2-x^2=(-x)^2=x^2\]

\[r^2=2x^2\]

\[\frac{r^2}{2}=x^2\]

\[\frac{\sqrt{r^2}}{\sqrt{2}}=y\]

Answer 6

\[\frac{\sqrt{2}\sqrt{r^2}}{2}\]

Answer 7

where r is your radius

Answer 8

sorry the last number was 2. i guess the radius is root of 2.