Question

The point 3,4 lies on the graph of f(x)= sqrt(2x+a). Which ordered pair below lies on the graph of g(x)=1/2sqrt(2x+a)?

Answer 1

$(3;\ 4);\ f(x)=\sqrt{2x+a}\\\\\sqrt{2\cdot6+a}=4\\\sqrt{12+a}=4\ \ \ /^2\\12+a=16\\a=16-12\\a=4\\\\f(x)=\sqrt{2x+4}$

$g(x)=\frac{1}{2}\sqrt{2x+4}\\\\a)\ \left(\frac{3}{2};\ 2\right)\\R=\frac{1}{2}\sqrt{2\cdot\frac{3}{2}+4}=\frac{1}{2}\sqrt{3+4}=\frac{1}{2}\sqrt7;\ L=2;\ L\neq R\\\\b)\ \left(\frac{5}{2};\ \frac{3}{2}\right)\\R=\frac{1}{2}\sqrt{2\cdot\frac{5}{2}+4}=\frac{1}{2}\sqrt{5+4}=\frac{1}{2}\sqrt9=\frac{1}{2}\cdot3=\frac{3}{2};\ L=\frac{3}{2};\ L=R$

$c)\ (3;\ 2)\\R=\frac{1}{2}\sqrt{2\cdot3+4}=\frac{1}{2}\sqrt{6+4}=\frac{1}{2}\sqrt{10};\ L=2;\ L\neq R\\\\d)\ (27;\ 8)\\R=\frac{1}{2}\sqrt{2\cdot27+4}=\frac{1}{2}\sqrt{54+4}=\frac{1}{2}\sqrt{58};\ L=8;\ L\neq R\\\\Answer:B$