Question

Create three function equations - one linear function, one quadratic function, and one cubicfunction — that all have a solution at (-5, 32).

Answer 1

Given point : (-5, 32)

A linear function would take the form:

`y\text{ = ax + b}`

Substituting the given point:

`-5a\text{ + b = 32}`

Our goal is to find the constants a and b.

set a = -5

`\begin{gathered} -5(-5)\text{ + b = 32} \\ 25\text{ + b = 32} \end{gathered}`

Solving for b:

`\begin{gathered} b\text{ = 32 - 25} \\ =\text{ 7} \end{gathered}`

Hence, the linear equation:

`y\text{ = -5x + 7}`

A quadratic function would take the form:

`y=ax^2+bx\text{ + c}`

Substituting the given point:

`\begin{gathered} (-5)^2a\text{ -5b + c = 32} \\ 25a\text{ -5b + c = 32} \end{gathered}`

Set a = 1 and b = -1:

`\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}`

Hence, the quadratic equation is:

`y=x^2\text{ -x + 2 }`

A cubic equation would take the form:

`y=ax^3+bx^2\text{ + cx + d}`

Substituting the given point:

`\begin{gathered} (-5)^3a+(-5)^2b\text{ -5c + d = 32} \\ -125a\text{ + 25b -5c + d = 32} \end{gathered}`

Set a = 1, b = 6, c = 1 :

`\begin{gathered} -125(1)\text{ + 25(6) - 5(1) + d = 32} \\ d\text{ = 32 -20} \\ d\text{ = 12} \end{gathered}`

Hence, the cubic equation would be:

`y=x^3+6x^2\text{ + x + 12}`