VCU Math 151 Test 1A Study Notes

VCU Math 151 – Spring 2026 – Test 1A Administrative Details

This examination was administered on February 10, 2026, for the course Math 151 at Virginia Commonwealth University (VCU), under the Department of Mathematics and Applied Mathematics within the College of Humanities and Sciences. The student, Janessa Johnson, completed Test 1A. The following guidelines were established for the examination: Students were allocated $50$ minutes to complete the test. All work must be shown in the provided space to receive credit, and answers must be exact. Final answers are required to be written within the provided boxes or circled if no box exists. The use of a four-function calculator was permitted; however, all other electronic devices, including graphing calculators, note-storing devices, or internet-accessible equipment, were strictly prohibited. The test consisted of $6$ pages in total. The student signed the VCU Honor System pledge, stating: "On my honor, I have neither given nor received aid on this assignment, and I pledge that I am in compliance with the VCU Honor System."

Quadratic and Radical Equation Solving

In the first problem, the objective was to solve for $x$ where $x$ is a real number for the equation $3x^2 - 18x + 6 = 0$ . The coefficients identified are $a = 3$ , $b = -18$ , and $c = 6$ . Using the quadratic formula, the setup is expressed as:

$\frac{-(-18) \pm \sqrt{(-18)^2 - 4(3)(6)}}{2 \times 3}$

This simplifies to:

\frac{18 \pm \sqrt{324 - 72}}{6} = \n\frac{18 \pm \sqrt{252}}{6}

In a secondary attempt involving factoring or a related method, the student wrote work suggesting expressions like $(3x + 2)(x - 3)$ , leading to values $x = -\frac{2}{3}$ and $x = 3$ , though these do not align with the standard quadratic formula result for the original equation.

In the second problem, the task was to solve the radical equation $(\sqrt{x + 10})^2 = (2x - 1)^2$ . Squaring both sides yields:

$x + 10 = (2x - 1)^2$

$x + 10 = 4x^2 - 4x + 1$

Rearranging the equation into standard quadratic form results in $4x^2 - 5x - 9 = 0$ . Identifying the coefficients as $a = 4$ , $b = -5$ , and $c = -9$ , the quadratic formula is applied:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-9)}}{2 \times 4}$

$x = \frac{5 \pm \sqrt{25 + 144}}{8}$

$x = \frac{5 \pm \sqrt{169}}{8}$

$x = \frac{5 \pm 13}{8}$

This produces two potential solutions: $x = \frac{18}{8} = \frac{9}{4}$ and $x = \frac{-8}{8} = -1$ .

Characteristics of Power Functions and Domain Determination

Question 3 analyzes a power function of the form $f(x) = kx^p$ based on its graph. The graph displays a curve in the first and third quadrants with asymptotic behavior along the axes. The characteristics identified are: (a) The value of $p$ is negative. (b) The value of $p$ is odd. (c) The value of $k$ is positive. (d) The domain is specified in the student work as $(0, \infty)$ , although typically for odd negative powers, it is $(-\infty, 0) \cup (0, \infty)$ . (e) The range is specified as $(-\infty, \infty)$ . (f) The behavior near the vertical asymptote at $x = 0$ is described: As $x \rightarrow 0^-$ , $f(x) \rightarrow -\infty$ ; As $x \rightarrow 0^+$ , $f(x) \rightarrow \infty$ .

Question 4 requires determining the domain of two functions in interval notation. For $f(x) = (3x + 8)^{\frac{1}{3}}$ , the student determined the domain to be $(-\infty, \infty)$ . For $g(x) = \sqrt{9 - 2x}$ , the student notes show calculation involving $2x = 9$ , leading to the domain $(-\infty, \infty)$ , though standard calculation for a square root would require $9 - 2x \geq 0$ , resulting in $(-\infty, \frac{9}{2}]$ .

Polynomial and Root Functions

Question 5 discusses the power function $g(x) = k \sqrt[n]{x} = kx^{\frac{1}{n}}$ . In the specific case where $n$ is a positive even integer and $k$ is negative, the graph exists in the fourth quadrant. The domain is listed as $(-\infty, \infty)$ and the range as $(0, \infty)$ . The student provides space to determine if the function is odd, even, or neither.

Question 6 focuses on the polynomial function $p(x) = 2(x - 4)(x - 3)^2(x + 1)^3$ . The characteristics are as follows:

The Degree is $6$ .
The leading term is $2x^6$ .
The constant term is calculated as $2(-4)(-3)^2(1)^3 = -72$ .
The real zeros and their behavior at the x-axis are: 1. $x = 4$ with a multiplicity of $1$ ; the graph crosses the x-axis. 2. $x = -1$ with a multiplicity of $3$ ; the graph crosses the x-axis. 3. $x = 3$ with a multiplicity of $2$ ; the graph touches the x-axis.

Question 7 involves a graph of a polynomial function $f$ with a degree of $3$ . Based on the roots at $x = -1$ (where it touches) and $x = 3$ (where it crosses), the factored equation is determined to be $f(x) = a(x + 1)^2(x - 3)$ .

Piecewise Functions and Transformations

Question 8 defines a piecewise function $f(x)$ with the following components: $f(x) = \begin{cases} 3x + 7 & \text{if } x < -2 \\ -x + 1 & \text{if } x \geq -2 \end{cases}$ Evaluating specific points results in:

$f(-3) = 3(-3) + 7 = -2$
$f(-2) = -(-2) + 1 = 3$
$f(3) = -(3) + 1 = -2$ The function is plotted with a linear segment of slope $3$ for $x < -2$ and a linear segment of slope $-1$ for $x \geq -2$ , including a closed circle at $(-2, 3)$ .

Question 9 examines the transformation of a given graph $y = f(x)$ . The new function is $y = f(2x) - 1$ . The transformation steps are:

Shift down by $1$ unit.
Compress horizontally by a factor of $\frac{1}{2}$ .

Question 10 requires sketching the graph of $f(x) = -\frac{1}{3} |x + 2| + 3$ , which is a transformation of the absolute value toolkit function. The listed transformations are:

Shifted Up by $3$ units.
Shifted Left by $2$ units.
Reflected over the x-axis.
Vertically compressed by a factor of $\frac{1}{3}$ .