📝 Graphing rational functions step by step (20 MCQs)
📖 From Calculus • 5. The derivative in Graphing and Applications • 20 questions available
What is Graphing rational functions step by step?
Definition:
Step-by-step graphing of rational functions involves identifying domain restrictions, intercepts, asymptotes (vertical, horizontal, slant), and using derivatives to find local extrema and concavity. This systematic process ensures all key features are captured for an accurate visual representation.
Example:
For , vertical asymptote at , horizontal at , decreasing on both intervals and .
Reason:
A structured approach prevents missing critical features like asymptotes or holes, which are essential for understanding the discontinuous nature of rational functions.
📝 All Graphing rational functions step by step MCQs
Q1. Which of the following best defines a rational function?
📖 Explanation: A rational function is precisely a quotient where both numerator and denominator are polynomials. This distinguishes it from other function types such as polynomial products or sums. Recognizing the ratio form is essential for later steps like locating asymptotes and simplifying the expression.
Q2. When does a rational function have a vertical asymptote?
📖 Explanation: A vertical asymptote occurs at any x‑value that makes the denominator zero provided that factor does not also appear in the numerator. If the factor cancels, the discontinuity becomes a removable hole rather than an asymptote. Thus the non‑cancelling zero of the denominator signals a vertical asymptote.
Q3. For f(x)=\\\frac{x-1}{x+2}\, what is the sign of f(x) on the interval \(-\\infty,-2)\?
📖 Explanation: When x is less than -2, both the numerator (x‑1) and denominator (x+2) are negative, producing a positive quotient. Evaluating a test point such as x= -3 confirms this: \(-3-1)/(-3+2)=(-4)/(-1)=4\, which is positive. Hence the function is positive throughout that interval.
Q4. For f(x)=\\\frac{x^{2}-4}{x-2}\, what is \\\displaystyle\\lim_{x\\to 2}f(x)\?
📖 Explanation: Factor the numerator: \x^{2}-4=(x-2)(x+2)\. Cancel the common factor (x‑2) to obtain \f(x)=x+2\ for x≠2. Taking the limit as x approaches 2 gives \2+2=4\. The limit exists even though the function is undefined at x=2, indicating a removable discontinuity.
Q5. What is the horizontal asymptote of \f(x)=\\frac{x^{2}}{x^{2}-9}\?
📖 Explanation: When the degrees of numerator and denominator are equal, the horizontal asymptote equals the ratio of the leading coefficients. Both numerator and denominator have leading coefficient 1, so the asymptote is \y=\\frac{1}{1}=1\. This line describes the end behavior as \x\\to\\pm\\infty\.
Q6. If the numerator degree is lower than the denominator degree, the end behavior of the rational function is that it
📖 Explanation: When the denominator’s degree exceeds the numerator’s, the denominator grows faster, forcing the quotient toward zero as \x\ becomes large in magnitude. This yields a horizontal asymptote at \y=0\. The other listed behaviors occur only when the numerator degree is equal to or greater than the denominator’s.
Q7. One of the vertical asymptotes of \f(x)=\\frac{x+1}{x^{2}-4}\ occurs at
📖 Explanation: The denominator factors as \(x-2)(x+2)\. Setting each factor to zero gives potential vertical asymptotes at \x=2\ and \x=-2\. Since neither factor cancels with the numerator, both are genuine asymptotes. Selecting one, \x=-2\ is a correct vertical asymptote.
Q8. For \f(x)=\\frac{2x-3}{x^{2}-9}\, which interval contains a point where \f(x)\ is positive?
📖 Explanation: Choose a test point, such as \x=4\. The numerator \2(4)-3=5\ is positive, and the denominator \4^{2}-9=7\ is also positive, giving a positive quotient. Therefore the interval \(3,\\infty)\ contains points where the function is positive.
Q9. What is \f(0)\ for \f(x)=\\frac{x-3}{x+3}\?
📖 Explanation: Substituting \x=0\ yields \f(0)=\\frac{0-3}{0+3}=\\frac{-3}{3}=-1\. The function is defined at \x=0\ because the denominator is non‑zero, so the value is a finite number, specifically \-1\.
Q10. Which statement about the end behavior of \f(x)=\\frac{3x^{2}+1}{2x^{2}-5}\ and \g(x)=\\frac{3x^{2}+1}{2x-5}\ is true?
📖 Explanation: For \f\, numerator and denominator have equal degree, giving a horizontal asymptote \y=\\frac{3}{2}\. For \g\, the numerator degree exceeds the denominator by one, producing an oblique (slant) asymptote rather than a horizontal one. Hence the second statement correctly describes their differing end behaviors.
Q11. Which rational function has a hole at \x=1\?
📖 Explanation: The expression \\\frac{x^{2}-1}{x-1}=\\frac{(x-1)(x+1)}{x-1}\ cancels the factor \(x-1)\, leaving \x+1\ with a removable discontinuity (hole) at \x=1\. The other choices either do not cancel or create a different type of discontinuity.
Q12. After simplifying \f(x)=\\frac{x^{3}-8}{x-2}\, what is the nature of the point \x=2\?
📖 Explanation: Factor the numerator: \x^{3}-8=(x-2)(x^{2}+2x+4)\. Cancel the common factor \(x-2)\ to obtain \f(x)=x^{2}+2x+4\ for \x\\neq2\. The original function is undefined at \x=2\, creating a removable discontinuity (hole) at that point.
Q13. Which function possesses a slant asymptote?
📖 Explanation: A slant (oblique) asymptote appears when the numerator degree exceeds the denominator degree by exactly one. In option C, \\\frac{x^{3}}{x^{2}+1}\ has numerator degree 3 and denominator degree 2, satisfying the condition and thus having a slant asymptote.
Q14. For \f(x)=\\frac{2x^{2}+3x-5}{x^{2}-4}\, at which \x\-value does the graph cross its horizontal asymptote?
📖 Explanation: The horizontal asymptote is \y=\\frac{2}{1}=2\. Setting \f(x)=2\ gives \2x^{2}+3x-5=2(x^{2}-4)\ → \2x^{2}+3x-5=2x^{2}-8\ → \3x-5=-8\ → \3x=-3\ → \x=-1\. Hence the graph meets its asymptote at \x=-1\.
Q15. Which statement about the symmetry of \f(x)=\\frac{x^{2}-9}{x^{2}+9}\ and \g(x)=\\frac{x^{2}-9}{x^{2}-9}\ is accurate?
📖 Explanation: Both functions involve only even powers of \x\; replacing \x\ with \-x\ leaves each expression unchanged, confirming even symmetry. Consequently each graph is symmetric about the y‑axis.
Q16. After simplifying \h(x)=\\frac{x^{3}-6x^{2}+9x}{x^{2}-3x}\, what is the nature of the point \x=0\?
📖 Explanation: Factor numerator as \x(x-3)^{2}\ and denominator as \x(x-3)\. Cancel the common factor \x\ to obtain \h(x)=x-3\ for \x\\neq0\. At \x=0\ the simplified expression equals \-3\, so the original function has a regular crossing point, not a discontinuity.
Q17. Which symmetry property indicates that \f(-x)=f(x)\?
📖 Explanation: The condition \f(-x)=f(x)\ defines even symmetry, meaning the graph is mirrored across the y‑axis. This property is characteristic of functions composed solely of even powers of \x\ or other even constructions.
Q18. Why does a rational function with numerator degree lower than denominator degree have horizontal asymptote \y=0\?
📖 Explanation: When the denominator’s polynomial degree exceeds that of the numerator, the denominator grows faster as \|x|\ becomes large. Consequently the quotient shrinks toward zero, producing a horizontal asymptote at \y=0\. The other statements describe different degree relationships.
Q19. Synthesize the steps to determine intervals of increase for \f(x)=\\frac{x^{2}-4}{x-1}\.
📖 Explanation: To locate increasing intervals, compute \f'(x)\ using the quotient rule, set the resulting numerator equal to zero to find critical points, and then test sign of \f'(x)\ in each interval determined by those points. This systematic approach reveals where the function rises.
Q20. Describe the behavior of \f(x)=\\frac{x^{2}+1}{x-2}\ as \x\ approaches 2 from the left and right.
📖 Explanation: The numerator \x^{2}+1\ is always positive. As \x\ approaches 2 from the left, the denominator \x-2\ is negative, so the quotient tends to \-\\infty\. Approaching from the right makes the denominator positive, sending the quotient to \+\\infty\. Thus