# Thread: check work 2

1. ## check work 2

1. the domain of $\displaystyle y= \frac{x-6}{x+6}$ is x cannot = -6

2. the function $\displaystyle f(x)=x^2(x-4)(x+3)^2(x+7)$ has 3 x intercepts

3. the x intercepts of $\displaystyle f(x)=x^2(2x-1)(3x+2)$ are x=1/2 and x=-2/3

2. Hello, William!

1. the domain of $\displaystyle y\:=\: \frac{x-6}{x+6}$ .is: .$\displaystyle x \neq -6$ . . . . Right!

2. The function $\displaystyle f(x)\:=\:x^2(x-4)(x+3)^2(x+7)$ .has 3 $\displaystyle x$-intercepts

3. The x-intercepts of $\displaystyle f(x)\:=\:x^2(2x-1)(3x+2)$ .are: .$\displaystyle x=\tfrac{1}{2}\,\text{ and }\,x=-\tfrac{2}{3}$
You're solving incorrectly . . .

(2) We have: .$\displaystyle x^2(x-4)(x+3)^2(x+7) \;=\;0$

Set each factor equal to zero and solve.

,$\displaystyle \begin{array}{cccccc}x^2 \:=\:0 & \Rightarrow & x \:=\:0 \\ x-4 \:=\:0 & \Rightarrow & x \:=\:4 \\ (x+3)^2 \:=\:0 & \Rightarrow & x \:=\:\text{-}3 \\ x+7 \:=\:0 & \Rightarrow & x \:=\:\text{-}7 \end{array} \quad\hdots\;four\: x\text{-intercepts}$