Answer:
[tex]W=\frac{2(x-2)}{(x-3)}[/tex]
Step-by-step explanation:
The correct question is
The area A of a rectangular garden is given by the expression 2x^2+2x− 12. The length L of the garden is given by the expression x^2 - 9. Find an expression for the width W of the garden. (Recall that A = LW). What are the excluded values in this calculation and what do they represent in context?
we know that
The area of a rectangular garden is given by the formula
[tex]A=LW[/tex]
we have
[tex]A=(2x^2+2x-12)\ units^2[/tex]
[tex]L=(x^2-9)\ units[/tex]
substitute
[tex](2x^2+2x-12)=(x^2-9)W[/tex]
Solve for W
[tex]W=\frac{(2x^2+2x-12)}{(x^2-9)}[/tex]
Find the roots of the quadratic equation of the numerator
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]2x^2+2x-12=0[/tex]
so
[tex]a=2\\b=2\\c=-12[/tex]
substitute in the formula
[tex]x=\frac{-2\pm\sqrt{2^{2}-4(2)(-12)}} {2(2)}[/tex]
[tex]x=\frac{-2\pm\sqrt{100}} {4}[/tex]
[tex]x=\frac{-2\pm10} {4}[/tex]
[tex]x=\frac{-2+10} {4}=2[/tex]
[tex]x=\frac{-2-10} {4}=-3[/tex]
so
The roots are x=2 and x=-3
The quadratic equation in factored form is equal to
[tex]2x^2+2x-12=2(x+3)(x-2)[/tex]
substitute in the above expression of W
[tex]W=\frac{2(x+3)(x-2)}{(x^2-9)}[/tex]
Rewrite the denominator as difference of squares
[tex](x^2-9)=(x+3)(x-3)[/tex]
substitute
[tex]W=\frac{2(x+3)(x-2)}{(x+3)(x-3)}[/tex]
Remember that
In a quotient, the denominator cannot be equal to zero
so
x=-3 and x=3 are excluded values
x=3 represent a vertical asymptote
x=-3 is not included in the domain of the function because the length cannot be a negative number
Simplify
[tex]W=\frac{2(x-2)}{(x-3)}[/tex]
