Respuesta :
Given point : (-5, 32)
A linear function would take the form:
[tex]y\text{ = ax + b}[/tex]Substituting the given point:
[tex]-5a\text{ + b = 32}[/tex]Our goal is to find the constants a and b.
set a = -5
[tex]\begin{gathered} -5(-5)\text{ + b = 32} \\ 25\text{ + b = 32} \end{gathered}[/tex]Solving for b:
[tex]\begin{gathered} b\text{ = 32 - 25} \\ =\text{ 7} \end{gathered}[/tex]Hence, the linear equation:
[tex]y\text{ = -5x + 7}[/tex]A quadratic function would take the form:
[tex]y=ax^2+bx\text{ + c}[/tex]Substituting the given point:
[tex]\begin{gathered} (-5)^2a\text{ -5b + c = 32} \\ 25a\text{ -5b + c = 32} \end{gathered}[/tex]Set a = 1 and b = -1:
[tex]\begin{gathered} 25(1)\text{ -5(-1) + c = 32} \\ 25\text{ + 5 + c = 32} \\ 30\text{ + c = 32} \\ Collect\text{ like terms:} \\ c\text{ = 32 - 30} \\ c\text{ = 2} \end{gathered}[/tex]Hence, the quadratic equation is:
[tex]y=x^2\text{ -x + 2 }[/tex]A cubic equation would take the form:
[tex]y=ax^3+bx^2\text{ + cx + d}[/tex]Substituting the given point:
[tex]\begin{gathered} (-5)^3a+(-5)^2b\text{ -5c + d = 32} \\ -125a\text{ + 25b -5c + d = 32} \end{gathered}[/tex]Set a = 1, b = 6, c = 1 :
[tex]\begin{gathered} -125(1)\text{ + 25(6) - 5(1) + d = 32} \\ d\text{ = 32 -20} \\ d\text{ = 12} \end{gathered}[/tex]Hence, the cubic equation would be:
[tex]y=x^3+6x^2\text{ + x + 12}[/tex]