Respuesta :

5) Let

[tex]y=ax^2+bx+c[/tex]

be the equation of the parabola that passes through the given points. Then we get that:

[tex]\begin{gathered} A)2=a(-2)^2+b(-2)+c=4a-2b+c, \\ B)8=a(1)^2+b\cdot1+c=a+b+c, \\ C)50=a(4)^2+b\cdot4+c=16a+4b+c. \end{gathered}[/tex]

Subtracting equation B) from equation A) and solving for a we get:

[tex]\begin{gathered} 2-8=4a-2b+c-a-b-c, \\ -6=3a-3b, \\ -2=a-b, \\ a=b-2. \end{gathered}[/tex]

Subtracting equation B) from equation C) and solving for a we get:

[tex]\begin{gathered} 50-8=16a+4b+c-a-b-c, \\ 42=15a+3b, \\ a=\frac{42-3b}{15}\text{.} \end{gathered}[/tex]

Therefore:

[tex]b-2=\frac{42-3b}{15}\text{.}[/tex]

Solving the above equation for b we get:

[tex]\begin{gathered} 15b-30=42-3b, \\ 18b=72, \\ b=4. \end{gathered}[/tex]

Substituting b=4 in a=b-2 we get:

[tex]a=4-2=2.[/tex]

Substituting a=2, b=4 in equation B), and solving for c we get:

[tex]\begin{gathered} 8=2+4+c, \\ c=2. \end{gathered}[/tex]

Answer: Second option.

6) Let

[tex]undefined[/tex]

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