Rational Equations and Inequalities Quiz

Reciprocal equation: $f(x) = \dfrac{1}{x}$. Property: Vertical asymptote at $x = 0$. Linear over linear equation: $f(x) = \dfrac{2x + 1}{3x - 2}$. Property: Horizontal asymptote at $y = \dfrac{2}{3}$. Quadratic over quadratic equation: $f(x) = \dfrac{x^2 - 4}{x^2 + 3x + 2}$. Property: Vertical asymptotes at $x = -2$ and $x = -1$. Quadratic over linear equation: $f(x) = \dfrac{x^2 - 1}{5x + 2}$. Property: Vertical asymptote at $x = -\dfrac{2}{5}$. Point of discontinuity does not exist for any of these equations.

The equation can be written as $f(x) = \dfrac{a(x - 3)}{(x - 3)(x + b)}$, where $a$ and $b$ are constants. To obtain a vertical asymptote at $x = 3$, we set the denominator equal to zero: $x - 3 = 0$, which gives $x = 3$. To find $b$, we set the numerator equal to zero at $x = 3$: $a(3 - 3) = 0$, which gives $a = 0$. Therefore, the equation is $f(x) = \dfrac{0}{(x - 3)(x + b)} = 0$.

To solve the inequality $f(x) > 3x^2 - 2x - 1$, we set $f(x)$ equal to the right-hand side of the inequality: $\dfrac{x - 1}{x + 2} > 3x^2 - 2x - 1$. We can multiply both sides of the inequality by $(x + 2)$ to eliminate the denominator: $(x - 1) > 3x^2 - 2x - 1(x + 2)$. Simplifying, we have $x - 1 > 3x^2 - 2x - x - 2$, which simplifies further to $x - 1 > 3x^2 - 3x - 2$. Rearranging, we get $3x^2 - 4x + 1 < 0$. To find the solution, we can factor the quadratic equation: $(x - 1)(3x - 1) < 0$. The critical points are $x = 1$ and $x = \dfrac{1}{3}$. Testing the intervals, we find that the solution is $x \in \left(\dfrac{1}{3}, 1\right)$. Graphing this solution on a number line, we have: $(-------------\bullet--\bullet------------)$.

<p>A reciprocal rational equation is of the form $f(x) = \frac{a}{x-b}$. The equation has a vertical asymptote at $x = b$ and a horizontal asymptote at $y = 0$. An example of a reciprocal rational equation is $f(x) = \frac{2}{x-1}$.</p> Signup and view all the answers

<p>A linear over linear rational equation is of the form $f(x) = \frac{ax+b}{cx+d}$. The equation has a vertical asymptote at $x = -d/c$ and a horizontal asymptote at $y = a/c$. An example of a linear over linear rational equation is $f(x) = \frac{2x+1}{3x-2}$.</p> Signup and view all the answers

<p>A quadratic over quadratic rational equation is of the form $f(x) = \frac{ax^2+bx+c}{dx^2+ex+f}$. The equation has vertical asymptotes determined by the roots of the denominator polynomial, and a horizontal asymptote determined by the leading coefficients of the numerator and denominator. An example of a quadratic over quadratic rational equation is $f(x) = \frac{x^2+1}{x^2-4}$.</p> Signup and view all the answers