How do you find the domain and range of $\frac{8 x - 48}{x^{2} - 13 x + 42}$ $(8x-48)/(x^2-13x+42)$ ?

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How do you find the domain and range of $\frac{8 x - 48}{x^{2} - 13 x + 42}$ $(8x-48)/(x^2-13x+42)$ ?

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Answer 1 · 2016-03-23T11:37:56

Domain (x) are all real numbers of x except $x = 7$ $x=7$
Range(y) are all real numbers except for $x = 7$ $x=7$

Explanation:

$y = \frac{8 x - 48}{x^{2} - 13 x + 42} = \frac{8 \cdot (x - 6)}{(x - 7) \cdot (x - 6)} = \frac{8}{x - 7}$ $y=(8x-48)/(x^2-13x+42) = (8*(x-6))/((x-7)*(x-6)) =8/(x-7)$
Domain (x) is all real numbers of x except $x = 7$ $x=7$ . As at x=7 denominator will be 0 and the function will not exist.
Range(y) is all real numbers except for $x = 7$ $x=7$

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How do you find the domain and range of $\frac{8 x - 48}{x^{2} - 13 x + 42}$ $(8x-48)/(x^2-13x+42)$ ?

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Explanation:

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How do you find the domain and range of 8x−48x2−13x+42(8x-48)/(x^2-13x+42)?

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Explanation:

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How do you find the domain and range of $\frac{8 x - 48}{x^{2} - 13 x + 42}$ $(8x-48)/(x^2-13x+42)$ ?