he function below. Then, identify the domain, range, and the location and type $0$ nuities. $\newline$ Domain: $\newline$ Range: $\newline$ Discontinuities: $\newline$ functions below, find $(g-f)(x)$ $\newline$ $8$ . $\newline$ Given the functions below, find and give its domain. its domain. $\newline$ $+2 x^{2}-7 x-1 ; g(x)=x^{3}-x^{2}+4 x \quad f(x)=\frac{2}{x+5} ; g(x)=\frac{1}{x}$

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Algebra 1

Domain and range of quadratic functions: equations

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Q. he function below. Then, identify the domain, range, and the location and type

0

nuities.

\newline

Domain:

\newline

Range:

\newline

Discontinuities:

\newline

functions below, find

(g-f)(x)

\newline

8

\newline

Given the functions below, find and give its domain. its domain.

\newline

+2 x^{2}-7 x-1 ; g(x)=x^{3}-x^{2}+4 x \quad f(x)=\frac{2}{x+5} ; g(x)=\frac{1}{x}

Identify Domain of $f(x)$ : Identify the domain of $f(x) = \frac{2}{(x+5)}$ . Since division by zero is undefined, $x+5$ cannot be zero. $x \neq -5$ . Domain of $f(x)$ : all real numbers except $x = -5$ .
Identify Range of $f(x)$ : Identify the range of $f(x) = \frac{2}{(x+5)}$ . As a rational function, $f(x)$ can take all real values except for the horizontal asymptote, if any. Since there's no restriction on the $y$ -values, the range is all real numbers.
Identify Discontinuities of $f(x)$ : Identify the discontinuities of $f(x) = \frac{2}{(x+5)}$ . Discontinuity occurs where the denominator is zero. $x+5 = 0$ leads to $x = -5$ . Type of discontinuity at $x = -5$ is a vertical asymptote.
Identify Domain of $g(x)$ : Identify the domain of $g(x) = \frac{1}{x}$ . $\newline$ Since division by zero is undefined, $x$ cannot be $0$ . $\newline$ Domain of $g(x)$ : all real numbers except $x = 0$ .
Identify Range of $g(x)$ : Identify the range of $g(x) = \frac{1}{x}$ . As a rational function, $g(x)$ can take all real values except for the horizontal asymptote, if any. Since there's no restriction on the $y$ -values, the range is all real numbers.
Identify Discontinuities of $g(x)$ : Identify the discontinuities of $g(x) = \frac{1}{x}$ . $\newline$ Discontinuity occurs where the denominator is zero. $\newline$ $x = 0$ leads to a vertical asymptote. $\newline$ Type of discontinuity at $x = 0$ is a vertical asymptote.
Calculate $(g-f)(x)$ : Calculate $(g-f)(x)$ for $g(x) = \frac{1}{x}$ and $f(x) = \frac{2}{x+5}$ .
$(g-f)(x) = g(x) - f(x) = \left(\frac{1}{x}\right) - \left(\frac{2}{x+5}\right)$ .
Find a common denominator and combine the fractions.
$(g-f)(x) = \frac{(x+5) - 2x}{x(x+5)}$ .
$(g-f)(x) = \frac{x+5 - 2x}{x^2+5x}$ .
$(g-f)(x) = \frac{-x+5}{x^2+5x}$ .
Identify Domain of $(g-f)(x)$ : Identify the domain of $(g-f)(x)$ . The domain is restricted by the denominators in the original functions. $x \neq 0$ and $x \neq -5$ . Domain of $(g-f)(x)$ : all real numbers except $x = 0$ and $x = -5$ .