Find all vertical asymptotes of the following function. f(x)=(x^(2)-16)/(2x^(2)+8x) Answer

Identify Vertical Asymptotes: To find the vertical asymptotes of the function, we need to determine where the denominator equals zero, as vertical asymptotes occur at values of x that make the denominator undefined. The denominator of the function is 2x^(2) + 8x. Set the denominator equal to zero and solve for x: 2x^(2) + 8x = 0Factor Out Common Factor: Factor out the common factor of 2x from the denominator: 2x(x + 4) = 0Solve for x: Set each factor equal to zero and solve for x: 2x = 0 or x + 4 = 0 This gives us x = 0 or x = -4.Check Numerator Values: However, we must check if these values of x also make the numerator zero, because if they do, they are not vertical asymptotes but rather holes in the graph. The numerator is x^(2) - 16, which factors to (x + 4)(x - 4). Substitute x = 0: (0 + 4)(0 - 4) = 16, which is not zero. Substitute x = -4: (-4 + 4)(-4 - 4) = 0, which is zero.Identify Vertical Asymptote: Since x = -4 makes both the numerator and the denominator zero, it is not a vertical asymptote but a hole in the graph. Therefore, the only vertical asymptote is at x = 0.

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Find all vertical asymptotes of the following function. $\newline$ $f(x)=\frac{x^{2}-16}{2x^{2}+8x}$ $\newline$ Answer

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Q. Find all vertical asymptotes of the following function.

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f(x)=\frac{x^{2}-16}{2x^{2}+8x}

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Answer

Identify Vertical Asymptotes: To find the vertical asymptotes of the function, we need to determine where the denominator equals zero, as vertical asymptotes occur at values of $x$ that make the denominator undefined. $\newline$ The denominator of the function is $2x^{2} + 8x$ . $\newline$ Set the denominator equal to zero and solve for $x$ : $\newline$ $2x^{2} + 8x = 0$
Factor Out Common Factor: Factor out the common factor of $2x$ from the denominator: $\newline$ $2x(x + 4) = 0$
Solve for x: Set each factor equal to zero and solve for x: $\newline$ $2x = 0$ or $x + 4 = 0$ $\newline$ This gives us $x = 0$ or $x = -4$ .
Check Numerator Values: However, we must check if these values of $x$ also make the numerator zero, because if they do, they are not vertical asymptotes but rather holes in the graph. $\newline$ The numerator is $x^{2} - 16$ , which factors to $(x + 4)(x - 4)$ . $\newline$ Substitute $x = 0$ : $(0 + 4)(0 - 4) = 16$ , which is not zero. $\newline$ Substitute $x = -4$ : $(-4 + 4)(-4 - 4) = 0$ , which is zero.
Identify Vertical Asymptote: Since $x = -4$ makes both the numerator and the denominator zero, it is not a vertical asymptote but a hole in the graph. $\newline$ Therefore, the only vertical asymptote is at $x = 0$ .