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zmey [24]
4 years ago
10

What is the domain and range of the relation shown in the table?

Mathematics
1 answer:
zysi [14]4 years ago
7 0
Domain is all the x values (2,-1,4,-3)and range is the y values (0,4,2,0)
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Dividing by a fraction is equivalent to multiply by its reciprocal, then:

\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}

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Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

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