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Math Help - check work 2

  1. #1
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    check work 2

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

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

    3. the x intercepts of f(x)=x^2(2x-1)(3x+2) are x=1/2 and x=-2/3
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  2. #2
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    Hello, William!

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

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

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

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

    Set each factor equal to zero and solve.

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

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