Mathematics: [tex]y+ rac{13}{4} = (x^2- rac{1}{2} )^2[/tex]

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[tex]y+ \frac{13}{4} = (x^2- \frac{1}{2} )^2[/tex]

Ответ:

kdelisa

10.03.2022

Mathematics

$\bold{y=\dfrac{1\pm \sqrt{4x+13}}{2}}$

Step-by-step explanation:

Inverse is when you swap the x's and y's and solve for y:

$y+\dfrac{13}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2\\\\\\x+\dfrac{13}{4}=\bigg(y-\dfrac{1}{2}\bigg)^2\rightarrow \text{(swapped the x's and y's)}\\\\\\\sqrt{x+\dfrac{13}{4}}=\sqrt{\bigg(y-\dfrac{1}{2}\bigg)^2}\rightarrow \text{(took square root of both sides)}\\\\\\\sqrt{\dfrac{4x+13}{4}}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(created common denominator in radical)}\\\\\\ \dfrac{\sqrt{4x+13}}{2}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(simplified radical)}$

$\pm \dfrac{\sqrt{4x+13}}{2}=y-\dfrac{1}{2}}\rightarrow \text{(divided both sides by}\ \pm )\\\\\\\dfrac{1}{2}\pm \dfrac{\sqrt{4x+13}}{2}=y\quad \rightarrow \text{(added}\ \dfrac{1}{2}\ \text{to both sides)}\\\\\\\dfrac{1\pm \sqrt{4x+13}}{2}=y\quad \rightarrow \text{(combined numerators)}$

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$y+ \frac{13}{4} = (x^2- \frac{1}{2} )^2$

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