Answer:
(5,0) & (3 , -1)
Step-by-step explanation:
Solution is the points where the parabola and line intersects.
Given:
The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).
To find:
The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.
Solution:
If a figure reflected across the line x=a, then
![(x,y)\to (-(x-a)+a,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%28-%28x-a%29%2Ba%2Cy%29)
![(x,y)\to (-x+a+a,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%28-x%2Ba%2Ba%2Cy%29)
![(x,y)\to (2a-x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%282a-x%2Cy%29)
The triangle after a reflection over x = -3. So, the rule of reflection is
![(x,y)\to (2(-3)-x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%282%28-3%29-x%2Cy%29)
![(x,y)\to (-6-x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Cto%20%28-6-x%2Cy%29)
The vertices of triangle after reflection are
![R(3,7)\to R'(-6-3,7)](https://tex.z-dn.net/?f=R%283%2C7%29%5Cto%20R%27%28-6-3%2C7%29)
![R(3,7)\to R'(-9,7)](https://tex.z-dn.net/?f=R%283%2C7%29%5Cto%20R%27%28-9%2C7%29)
Similarly,
![S(-5,-2)\to S'(-6-(-5),-2)](https://tex.z-dn.net/?f=S%28-5%2C-2%29%5Cto%20S%27%28-6-%28-5%29%2C-2%29)
![S(-5,-2)\to S'(-6+5,-2)](https://tex.z-dn.net/?f=S%28-5%2C-2%29%5Cto%20S%27%28-6%2B5%2C-2%29)
![S(-5,-2)\to S'(-1,-2)](https://tex.z-dn.net/?f=S%28-5%2C-2%29%5Cto%20S%27%28-1%2C-2%29)
And,
![T(3,-5)\to T'(-6-3,-5)](https://tex.z-dn.net/?f=T%283%2C-5%29%5Cto%20T%27%28-6-3%2C-5%29)
![T(3,-5)\to T'(-9,-5)](https://tex.z-dn.net/?f=T%283%2C-5%29%5Cto%20T%27%28-9%2C-5%29)
Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).
I think it is C sorry if i am wrong but i am confident
Foundry is what you get when you smelt iron :)