First you would need to combine like terms. do you know how to do that?
well, if that function f(x) were to be continuos on all subfunctions, that means that whatever value 7x + k has when x = 2, meets or matches the value that kx² - 6 has when x = 2 as well, so then 7x + k = kx² - 6 when f(2)
![f(x)= \begin{cases} 7x+k,&x\leqslant 2\\ kx^2-6&x > 2 \end{cases}\qquad \qquad f(2)= \begin{cases} 7(2)+k,&x\leqslant 2\\ k(2)^2-6&x > 2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ 7(2)+k~~ = ~~k(2)^2-6\implies 14+k~~ = ~~4k-6 \\\\\\ 14~~ = ~~3k-6\implies 20~~ = ~~3k\implies \cfrac{20}{3}=k](https://tex.z-dn.net/?f=f%28x%29%3D%20%5Cbegin%7Bcases%7D%207x%2Bk%2C%26x%5Cleqslant%202%5C%5C%20kx%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20f%282%29%3D%20%5Cbegin%7Bcases%7D%207%282%29%2Bk%2C%26x%5Cleqslant%202%5C%5C%20k%282%29%5E2-6%26x%20%3E%202%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207%282%29%2Bk~~%20%3D%20~~k%282%29%5E2-6%5Cimplies%2014%2Bk~~%20%3D%20~~4k-6%20%5C%5C%5C%5C%5C%5C%2014~~%20%3D%20~~3k-6%5Cimplies%2020~~%20%3D%20~~3k%5Cimplies%20%5Ccfrac%7B20%7D%7B3%7D%3Dk)
The equation of a line is y-mx+b with m being slope and b being y-intercept. Using the values provided in your problem, your answer would be y=2/3x+9, or C.
Answer: 1 is not a perfect square. 3 is the only prime number one less than a square.