Question

May someone help me solve d and e.

Answer 1

Answer:

d.

$x = 2 {y}^{2} - 4 \\ 2 {y}^{2} = x + 4 \\ \\ {y}^{2} = \frac{x + 4}{2} \\ \\ y = \frac{ + }{} \sqrt{ \frac{x + 4}{2} } \\ \\ y = \frac{ + }{} \frac{ \sqrt{x + 4} }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } \\ \\ y = \frac{ + }{} \: \frac{ \sqrt{x + 4} \times \sqrt{2} }{ \sqrt{2} \sqrt{2} } \\ \\ y = \frac{ + }{} \frac{ \sqrt{2(x + 4)} }{2} \\ \\ y =\frac{ \sqrt{2(x + 4)} }{2} \\ \\ y = - \frac{ \sqrt{2(x + 4)} }{2}$

e.

$x = {(y - 5)}^{2}$

with the root of both sides

$\sqrt{x} = \sqrt{ {(y - 5)}^{2} } \\ \\ \frac{ + }{} \sqrt{x} = y - 5 \\ \\ y = \sqrt{x} + 5 \\ y = - \sqrt{x} + 5$

I hope I helped you^_^