Math 112 - Practice Exam 1 Notes
General Information
Exam Type: Precalculus
Duration: 50-minute
Total Points: 100 (including 10 bonus points possible)
Key topics covered in Precalculus relevant to this exam include linear equations, quadratic equations, circles, parabolas, and fundamental concepts of functions, all of which are essential building blocks for calculus.
Problem 1: Unit Circle Equation
Points: 20 points
Task: Find the equation of a unit circle with center at the intersection of two given lines.
Line 1: y_1 = -x + 1
Line 2: y_2 = x - 1
The unit circle is defined as a circle with a radius of r = 1 . The center of this circle is the point where the two given lines intersect, as specified in the problem statement.
Solution Steps:
Solve for the intersection point of the two lines: To find the point where the lines intersect, we set their equations equal to each other, as both y1 and y2 represent the same y-coordinate at the intersection.
Set the equations equal: -x + 1 = x - 1
Simplifying the equation by adding x to both sides and adding 1 to both sides yields: 2 = 2x , which implies x = 1 .
Substitute the value of x = 1 back into either of the original line equations (e.g., y = x - 1 ) to find the corresponding y-coordinate: y = 1 - 1 = 0 .
Intersection point: (1, 0). This point will serve as the center of our unit circle.
Determine the equation of the unit circle: The standard form for the equation of a circle is (x - h)^2 + (y - k)^2 = r^2 , where (h, k) is the center and r is the radius.
Center: (h, k) = (1, 0) (from the intersection point).
Radius: r = 1 (as it's a unit circle).
Substitute these values into the standard formula:
Final equation: (x - 1)^2 + (y - 0)^2 = 1^2 which simplifies to (x - 1)^2 + y^2 = 1 .
Graphical Representation: The visualization confirms the algebraic solution.
Coordinates on the graph depict the intersection at (1, 0), which is the center of the circle.
The graph displays the unit circle itself, centered precisely at (1, 0).
Graphical Elements
Coordinate axes: The x-axis (horizontal) ranges from -1.5 to 2.5, and the y-axis (vertical) ranges from -1.5 to 1.5, providing a clear view of the circle and its center.
Labeling: The intersection point (1, 0) is distinctly marked with a filled red circle to highlight the center of the unit circle.
Problem 2: Line Equations and Sketches
Points: 30 points total. This problem assesses the understanding of linear equations, slope, and relationships between lines (parallel and perpendicular).
Breakdown of tasks:
Part (a): Line through P1 and P2
Given Points: P1 = (-1, -1) and P2 = (0, 0).
Slope Calculation: The slope mm of a line passing through two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is given by the formula: m = rac{y2 - y1}{x2 - x1} . The slope represents the rate of change of yy with respect to xx.
Applying the formula with P1 and P2: m = rac{0 - (-1)}{0 - (-1)} = rac{1}{1} = 1 . The positive slope indicates an upward trend from left to right.
Equation Derivation: Using the point-slope form, y - y1 = m(x - x1) , which is a highly versatile method for finding the equation of a line when a point and the slope are known. Using point P2 = (0, 0) simplifies the calculation significantly:
y - 0 = 1(x - 0)
Results in: y = x . This is the equation of the line passing through points P1 and P2.
Intercepts: For the line y = x , both the x-intercept and the y-intercept are at (0, 0).
Quicker Solution Insight: Observing that for both given points, the x-coordinate is equal to the y-coordinate directly indicates that the relationship y = x holds true for the line.
Part (b): Line parallel to Part (a) through P3
Given Point: P3 = (0, 1).
Equation Construction: Parallel lines have identical slopes. Therefore, the slope of this new line, m_{par} , will be the same as the slope from part (a).
Slope (same as part a): m_{par} = 1 .
We can use the slope-intercept form, y = mx + b , where m is the slope and b is the y-intercept. Since we know the slope and a point (P3), we can substitute these values to find b .
From point P3 = (0, 1), substitute x=0 and y=1 into the equation: 1 = 1(0) + b . Hence, b = 1 .
Final Equation: y = x + 1 . This line is shifted one unit up from the line in Part (a).
Quick Insight: When a line passes through the y-axis at (0, b), the value of b is directly the y-intercept. For P3 = (0, 1), the y-coordinate directly gives b = 1 .
Part (c): Line perpendicular to Part (a)
Through Point P2: For two non-vertical lines to be perpendicular, their slopes must be negative reciprocals of each other. If the original slope is m , the perpendicular slope, m_{perp} , is - rac{1}{m} . The line must pass through P2 = (0, 0).
Perpendicular slope: Given m = 1 from part (a), the negative reciprocal is m_{perp} = - rac{1}{1} = -1 .
Point-slope form: Using point P2 = (0, 0) and the perpendicular slope:
y - 0 = -1(x - 0)
Results in: y = -x . This line has a negative slope, indicating a downward trend.
Quick Insight: This line reflects the relationship where y = -x , implying that for any point on the line, the y-coordinate is the negation of the x-coordinate. The perpendicular relationship gives the inverse slope directly, leading to immediate understanding of characteristics.
Problem 3: Parabola Through Points
Bonus Points: 10 points. This problem tests the ability to determine the equation of a parabola given its vertex and another point.
Given Point and Vertex:
Vertex: P3 = (0, 1). When the vertex of a parabola is on the y-axis (i.e., its x-coordinate is 0), the general quadratic equation simplifies from f(x) = a(x - h)^2 + k to f(x) = a(x - 0)^2 + k , or simply f(x) = ax^2 + k . Here, (h, k) = (0, 1) , so k = 1 .
The parabola must also pass through point P1 = (-1, -1). This point will be used to find the leading coefficient a .
Quadratic Equation Structure:
General form for this parabola: f(x) = ax^2 + k , where k = 1 is the y-coordinate of the vertex.
Using point P1 = (-1, -1) to find a : Substitute x = -1 and f(x) = -1 into the equation:
f(-1) = a(-1)^2 + 1 = a(1) + 1 = a + 1 .
Since f(-1) = -1 , we set: a + 1 = -1
Results in: a = -2 .
Final Equation: Substituting a = -2 and k = 1 back into the general form:
f(x) = -2x^2 + 1 . The negative value of a = -2 indicates that the parabola opens downwards, signifying that the vertex is a maximum point.
Graphical Representation:
The graph illustrates the parabola f(x) = -2x^2 + 1 , showing that it passes through the specified points P1 = (-1, -1), P2 = (0, 0), and P3 = (0, 1), with P3 being its vertex. This visual representation helps confirm the accuracy of the derived equation.
Problem 4: Bacterial Population Growth Rate Modeling
Points: 50 points total. This problem delves into the analysis of a quadratic function in the context of population growth, covering various forms and interpretations.
Given model for the rate of population growth: r(p) = p(P - p) .
Variables defined:
p > 0 : represents the current population size. It must be positive as it refers to a number of bacteria.
P : represents the maximum possible population size that the environment can sustain (carrying capacity).
Part (a): Form Identification
Form Type: The rate function r(p) = p(P - p) is in factored form. This form is particularly useful for quickly identifying the roots (or x-intercepts) of the quadratic equation, which are the points where the rate of growth is zero. In this case, the roots are when p = 0 or P - p = 0 \implies p = P . These roots signify population sizes where the growth stops or hasn't started.
Part (b): Standard Form Calculation
To convert to standard form, ax^2 + bx + c , we need to expand the factored expression.
Expand the function:
r(p) = p(P - p) = Pp - p^2 .
Standard form result: By rearranging the terms in descending order of powers of p , we get r(p) = -p^2 + Pp . (Or -1p^2 + Pp + 0)
Leading Coefficient: The coefficient of the p^2 term is -1 . A negative leading coefficient indicates that the parabola opens downward, which means the function r(p) will have a maximum value. In the context of population growth rate, this maximum represents the optimal population size at which growth is fastest.
Part (c): Vertex Form Conversion
To convert the standard form r(p) = -p^2 + Pp to vertex form, a(p - h)^2 + k , we use the method of completing the square.
Completing the Square Derivation:
Start with: r(p) = -p^2 + Pp
Factor out the leading coefficient (in this case, -1) from the terms involving p :
r(p) = -(p^2 - Pp)
To complete the square inside the parenthesis, take half of the coefficient of p (which is -P ), square it (which is (P/2)^2 = P^2/4 ), and add and subtract it inside the parenthesis:
r(p) = -(p^2 - Pp + rac{P^2}{4} - rac{P^2}{4})
Now, group the perfect square trinomial:
r(p) = -((p - rac{P}{2})^2 - rac{P^2}{4})
Distribute the negative sign back into the terms:
r(p) = -(p - rac{P}{2})^2 + rac{P^2}{4} .
Vertex Identification: From the vertex form r(p) = -(p - rac{P}{2})^2 + rac{P^2}{4} , the vertex coordinates (h, k) are directly identifiable.
Vertex Coordinates: (h, k) = ( rac{P}{2}, rac{P^2}{4}) . Since the parabola opens downward (as identified in Part b), this vertex represents a global maximum for the rate of population growth. This means the population grows fastest when p = rac{P}{2} , i.e., when the population is half of the maximum possible population.
Part (d): Domain and Range Identification
Domain Consideration: The domain defines the possible values for the current population p . Realistically, a population must be positive, and it cannot exceed the maximum possible population P . If p = 0 , there is no population to grow, and if p = P , the population has reached its maximum, and the growth rate becomes zero due to limiting resources.
Therefore, the domain for p is 0 < p \le P .
Range Results: The range defines the possible values for the rate of growth r(p) . The growth rate must be non-negative (populations don't shrink by growing at a negative rate in this model, they just stop growing or don't grow). The minimum rate occurs at the roots of the function ( p=0 and p=P ), where r(p) = 0 .
The maximum rate occurs at the vertex, where p = rac{P}{2} , yielding the maximum value for r(p) = rac{P^2}{4} .
Thus, the range for the rate of growth is: 0 \le r(p) \le rac{P^2}{4} .
Part (e): Graph Sketching
Sketch Requirements: A well-labeled sketch provides a visual summary of the function's behavior.
Draw a downward-opening parabola, consistent with the negative leading coefficient.
Explicitly label the roots (x-intercepts) at p = 0 and p = P , where the growth rate is zero.
Mark the vertex at its coordinates ( rac{P}{2}, rac{P^2}{4}) . This point should clearly represent the peak of the parabola, indicating the maximum rate of population growth.
Graphical Elements for Rate of Population Growth (Fig. 3)
The figure illustrates the function r(p) = -p^2 + Pp for various maximum population values like P = 1, 2, and 3. This demonstrates how the shape and peak of the parabola change with different carrying capacities ( P ).
Coordinate values are clearly indicated, illustrating the defined relationships and derived equations by showing the positions of the roots and the vertex for each P value. This helps in understanding the model's implications for different biological contexts.