MATH 126 Quiz 5 Solutions

Vocabulary practice flashcards covering all problem topics from MATH 126 Quiz 5 Solutions, including linear intercepts, rationalization, rational expressions, quadratic equations, function evaluation, difference quotients, piecewise functions, domains, and graphical behaviors.

$x$-intercept of $y = 5x - 3$

The point $(\frac{3}{5}, 0)$, where the graph crosses the $x$-axis.

$y$-intercept of $y = 5x - 3$

The point $(0, -3)$, where the graph crosses the $y$-axis.

Rationalized Form of $\frac{2}{2 - \sqrt{5}}$

$-\frac{2(2 + \sqrt{5})}{3}$, obtained by multiplying the numerator and denominator by the conjugate of the denominator.

Simplified Form of $\frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}$

$\frac{x^2 + x + 1}{x^3}$, obtained by expressing each fraction over the common denominator $x^3$.

Solutions to $4x^2 + 7x - 2 = 0$

$x = \frac{1}{4}$ and $x = -2$, the roots found by solving the quadratic equation.

Value of $f(-1)$ for $f(x) = x^2 + \frac{1}{x}$

$0$, evaluated as $(-1)^2 + \frac{1}{-1} = 1 - 1 = 0$.

Value of $f(0)$ for $f(x) = x^2 + \frac{1}{x}$

$\text{Undefined}$, because substituting $x = 0$ results in division by zero: $(0)^2 + \frac{1}{0}$.

Value of $f(1)$ for $f(x) = x^2 + \frac{1}{x}$

$2$, evaluated as $(1)^2 + \frac{1}{1} = 1 + 1 = 2$.

Expression for $f(a)$ given $f(x) = 2x^2 - 3$

$2a^2 - 3$, obtained by replacing the variable $x$ with $a$.

Expression for $f(a+h)$ given $f(x) = 2x^2 - 3$

$2a^2 + 4ah + 2h^2 - 3$, obtained by expanding $2(a+h)^2 - 3$.

Difference Quotient for $f(x) = 2x^2 - 3$

$\frac{f(a+h) - f(a)}{h} = 4a + 2h$, simplified through substitution, combining like terms, factoring out $h$, and canceling common terms.

Value of $f(-4)$ for the Piecewise Function $f(x)$

$20$, evaluated using the rule $x^2 - x$ for the interval $x < -3$: $(-4)^2 - (-4) = 20$.

Value of $f(-3)$ for the Piecewise Function $f(x)$

$-2$, evaluated using the rule $x + 1$ for the interval $-3 \le x < 1$: $(-3) + 1 = -2$.

Value of $f(0)$ for the Piecewise Function $f(x)$

$1$, evaluated using the rule $x + 1$ for the interval $-3 \le x < 1$: $(0) + 1 = 1$.

Value of $f(1)$ for the Piecewise Function $f(x)$

$5$, evaluated using the constant rule $5$ for the interval $x \ge 1$.

Domain of $f(x) = x^2 + 5x + 6$

$(-\infty, \infty)$; because the function is a polynomial, there are no restrictions on $x$.

Domain of $f(x) = \frac{x}{x+1}$

$(-\infty, -1) \cup (-1, \infty)$; the denominator is not allowed to be zero, excluding $x = -1$.

Domain of $f(x) = \sqrt{x^2 - 4}$

$(-\infty, -2] \cup [2, \infty)$; the expression inside the radical must be non-negative ($x^2 - 4 \ge 0$).

Domain of $f(x) = \sqrt{x^2 + 4}$

$(-\infty, \infty)$; the radicand must be non-negative, and $x^2 + 4$ is positive for all real numbers.

Domain of $f(x) = \frac{\sqrt{x-3}}{x^2 + 5x + 6}$

$[3, \infty)$; restricted by the numerator requiring $x \ge 3$ and the denominator requiring $x \ne -2, -3$.

Domain of $f(x) = \frac{\sqrt{x+5}}{x^2 + 5x + 6}$

$[-5, -3) \cup (-3, -2) \cup (-2, \infty)$; restricted by the numerator requiring $x \ge -5$ and the denominator excluding $x = -3$ and $x = -2$.

Domain (from Graph)

$[-3, \infty)$, representing all input $x$-values included on the graph.

Range (from Graph)

$(-\infty, 4]$, representing all output $y$-values achieved by the graph.

Increasing Domain (from Graph)

$(-1, 2)$, the interval of $x$-values where the function is rising from left to right.

Decreasing Domain (from Graph)

$(-3, -1), (2, \infty)$, the intervals of $x$-values where the function is falling from left to right.

Constant Domain (from Graph)

$\text{None}$, indicating there are no intervals where the graph remains horizontal.

Local Minimums (from Graph)

The point $(-1, -1)$, where the graph achieves a local low point.

Local Maximums (from Graph)

The point $(2, 4)$, where the graph achieves a local high point.