Answer:
what do you need? ^-^ I can help you
Answer:
because I said so....that's why
so we have a table of values, with x,y coordinates, so let's use any two of those points to get the slope of the table and use the point-slope form to get its equation
![~\hspace{2.7em}\stackrel{\textit{let's use}}{\downarrow }\qquad \stackrel{\textit{and this}}{\downarrow }\\\begin{array}{|lr|r|r|r|r|}\cline{1-6}x&0&1&2&3&4\\y&-1&3&7&11&15\\\cline{1-6}\end{array}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad(\stackrel{x_2}{4}~,~\stackrel{y_2}{15})](https://tex.z-dn.net/?f=~%5Chspace%7B2.7em%7D%5Cstackrel%7B%5Ctextit%7Blet%27s%20use%7D%7D%7B%5Cdownarrow%20%7D%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Band%20this%7D%7D%7B%5Cdownarrow%20%7D%5C%5C%5Cbegin%7Barray%7D%7B%7Clr%7Cr%7Cr%7Cr%7Cr%7C%7D%5Ccline%7B1-6%7Dx%260%261%262%263%264%5C%5Cy%26-1%263%267%2611%2615%5C%5C%5Ccline%7B1-6%7D%5Cend%7Barray%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B15%7D%29)

The A’ is your reflected shape. The red line is y=-1, or your line right reflection. To solve, I counted the distance from the red line on the original triangle, and then the same on the opposite side of the red line, and counted the same distance, plotting the points at the final spot.
Answer:
Step-by-step explanation:
y + 2 = 7(x - 2)
y + 2 = 7x + 14
y = 7x + 12