May someone help me solve d and e.
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Answer:
d.
[tex]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}[/tex]
e.
[tex]x = {(y - 5)}^{2} [/tex]
with the root of both sides
[tex] \sqrt{x} = \sqrt{ {(y - 5)}^{2} } \\ \\ \frac{ + }{} \sqrt{x} = y - 5 \\ \\ y = \sqrt{x} + 5 \\ y = - \sqrt{x} + 5[/tex]
I hope I helped you^_^