{"id":182361,"date":"2025-01-13T17:46:06","date_gmt":"2025-01-13T17:46:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182361"},"modified":"2025-01-13T17:46:08","modified_gmt":"2025-01-13T17:46:08","slug":"find-the-horizontal-intercepts-the-vertical-intercept","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/find-the-horizontal-intercepts-the-vertical-intercept\/","title":{"rendered":"Find the horizontal intercepts, the vertical intercept"},"content":{"rendered":"\n

Find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the function. Use that information to sketch a graph.<\/p>\n\n\n\n

f(x)=x^2\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122x\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u212220\/x^2\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u212216<\/p>\n\n\n\n

Enter the intercepts as points,\u00c3\u201a\u00c2 (a,b).<\/p>\n\n\n\n

The x-intercept is_____________<\/p>\n\n\n\n

The y-intercept is__________<\/p>\n\n\n\n

The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;6 or x+1;x\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u21221). The order of the list does not matter.<\/p>\n\n\n\n

Vertical asymptotes:<\/p>\n\n\n\n

x=<\/p>\n\n\n\n

Horizontal or slant asymptote:<\/p>\n\n\n\n

y=<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To analyze the function \ue203f(x)=x2\u2212x\u221220×2\u221216f(x) = \\frac{x^2 – x – 20}{x^2 – 16}\ue204, we’ll determine its intercepts and asymptotes, and then sketch its graph.<\/p>\n\n\n\n

1. Vertical Asymptotes:<\/strong><\/p>\n\n\n\n

\ue203Vertical asymptotes occur where the denominator is zero, provided the numerator isn’t zero at those points.\ue204\ue206<\/p>\n\n\n\n

\ue203The denominator is x2\u221216=(x\u22124)(x+4)x^2 – 16 = (x – 4)(x + 4). Setting it to zero gives:\ue204\ue206<\/p>\n\n\n\n

\ue203x2\u221216=0\u2005\u200a\u27f9\u2005\u200ax=\u00b14x^2 – 16 = 0 \\implies x = \\pm 4\ue204\ue206<\/p>\n\n\n\n

\ue203At x=4x = 4:\ue204\ue206<\/p>\n\n\n\n

\ue203f(4)=42\u22124\u22122042\u221216=16\u22124\u22122016\u221216=\u221280\u2005\u200a\u27f9\u2005\u200aundefinedf(4) = \\frac{4^2 – 4 – 20}{4^2 – 16} = \\frac{16 – 4 – 20}{16 – 16} = \\frac{-8}{0} \\implies \\text{undefined}\ue204\ue206<\/p>\n\n\n\n

\ue203At x=\u22124x = -4:\ue204\ue206<\/p>\n\n\n\n

\ue203f(\u22124)=(\u22124)2+4\u221220(\u22124)2\u221216=16+4\u22122016\u221216=00\u2005\u200a\u27f9\u2005\u200aindeterminatef(-4) = \\frac{(-4)^2 + 4 – 20}{(-4)^2 – 16} = \\frac{16 + 4 – 20}{16 – 16} = \\frac{0}{0} \\implies \\text{indeterminate}\ue204\ue206<\/p>\n\n\n\n

\ue203Since both points lead to division by zero, x=4x = 4 and x=\u22124x = -4 are vertical asymptotes.\ue204\ue206<\/p>\n\n\n\n

2. Horizontal Asymptote:<\/strong><\/p>\n\n\n\n

\ue203For rational functions, if the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of their leading coefficients.\ue204\ue206<\/p>\n\n\n\n

\ue203Here, both numerator and denominator are degree 2, with leading coefficients 1. Thus, the horizontal asymptote is:\ue204\ue206<\/p>\n\n\n\n

\ue203y=11=1y = \\frac{1}{1} = 1\ue204\ue206<\/p>\n\n\n\n

3. Intercepts:<\/strong><\/p>\n\n\n\n