Answer:
[tex]\textsf{A. } y=3x^2+10x-8[/tex]
Step-by-step explanation:
We are given the information that a function has zeros at x = ⅔ and x = -4. In order to find the function that has those zeros, we can substitute the value of the zeros into each function. If the value of a function equates to zero with both values, then that is the function we are looking for.
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Standard Form of a Quadratic: ax² + bx + c = 0.
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[tex]\large \text{$y = 3x^2+10x-8 \implies 0=3x^2+10x-8$}[/tex]
[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=3\left(\dfrac{2}{3}\right)^2+10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=3\left(\dfrac{4}{9}\right)+10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=\not{3}\left(\dfrac{4}{\not{9}\ 3}\right)+\dfrac{20}{3}-8\\\\\implies 0=\dfrac{4}{3}+\dfrac{20}{3}-8\\\\\implies 0=\dfrac{24}{3}-8\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex] [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=3(-4)^2+10(-4)-8\\\\\implies 0=3(16)-40-8\\\\\implies 0=48-48\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex]
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[tex]\large \text{$y = 2x^2-5x-12 \implies 0=2x^2-5x-12$}[/tex]
[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=2\left(\dfrac{2}{3}\right)^2-5\left(\dfrac{2}{3}\right)-12\\\\\implies 0=2\left(\dfrac{4}{9}\right)-\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}-\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}-\dfrac{10\times3}{3\times3}-\dfrac{12\times9}{9}\\\\\implies 0=\dfrac{8}{9}-\dfrac{30}{9}-\dfrac{108}{9}\\\\\implies0=-\dfrac{22}{9}-\dfrac{108}{9}\\\\\implies 0=-\dfrac{130}{9}\ \textsf{X}\end {minipage}}[/tex] [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=2(-4)^2-5(-4)-12\\\\\implies 0=2(16)+20-12\\\\\implies 0=32+20-12\\\\\implies 0=40\ \textsf{X}\end{minipage}}[/tex]
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[tex]\large \text{$y = 2x^2+5x-12 \implies 0=2x^2+5x-12$}[/tex]
[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=2\left(\dfrac{2}{3}\right)^2+5\left(\dfrac{2}{3}\right)-12\\\\\implies 0=\dfrac{8}{9}\right)+\dfrac{10}{3}-12\\\\\implies 0=\dfrac{8}{9}\right)+\dfrac{10\times3}{3\times3}-\dfrac{12\times9}{9}\\\\\implies 0=\dfrac{8}{9}+\dfrac{30}{9}-\dfrac{108}{9}\\\\\implies0=\dfrac{38}{9}-\dfrac{108}{9}\\\\\implies 0=-\dfrac{70}{9}\ \textsf{X}\end{minipage}}[/tex] [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=2(-4)^2+5(-4)-12\\\\\implies 0=2(16)-20-12\\\\\implies 0=32-32\\\\\implies 0=0\ \checkmark\end{minipage}}[/tex]
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[tex]\large \text{$y = 3x^2-10x-8 \implies 0=3x^2-10x-8$}[/tex]
[tex]\boxed{\begin{minipage}{15 em}{\text{$x=\dfrac{2}{3}$}} \\ \\\implies 0=3\left(\dfrac{2}{3}\right)^2-10\left(\dfrac{2}{3}\right)-8\\\\\implies 0=3\left(\dfrac{4}{9}\right)-\dfrac{20}{3}-8\\\\\implies 0=\not{3}\left(\dfrac{4}{\not{9}\ 3}\right)-\dfrac{20}{3}-8\\\\\implies 0=\dfrac{4}{3}-\dfrac{20}{3}-\dfrac{8\times3}{3}\\\\\implies 0=-\dfrac{16}{3}-\dfrac{24}{3}\\\\\implies 0=-\dfrac{40}{3}\ \textsf{X}\end{minipage}}[/tex] [tex]\boxed{\begin{minipage}{15 em}{\text{$x=-4$}} \\ \\\implies 0=3(-4)^2-10(-4)-8\\\\\implies 0=3(16)+40-8\\\\\implies 0=48+40-8\\\\\implies 0=80\ \textsf{X}\end{minipage}}[/tex]
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Therefore, the function whose zeros are ⅔ and -4 is [tex]y=3x^2+10x-8[/tex].