Precalculus Practice Test 1 – Step-by-Step Study Guidance

Q1(a). Find the equation of the line that passes through the point (1, -3) and is parallel to the line 2x - y = 16.

Background

Topic: Equations of Lines

This question tests your understanding of how to find the equation of a line given a point and a condition of parallelism to another line.

Key Terms and Formulas

• Slope-intercept form: $y = mx + b$

• Parallel lines: Have the same slope.

• Standard form: $Ax + By = C$

Step-by-Step Guidance

1. Rewrite the given line $2x - y = 16$ in slope-intercept form to identify its slope.

2. Recall that parallel lines have the same slope, so use this slope for your new line.

3. Use the point-slope form $y - y_1 = m(x - x_1)$ with the point (1, -3) and the slope found in step 1.

4. Simplify the equation to your preferred form (slope-intercept or standard form), but stop before plugging in all values or simplifying completely.

Try solving on your own before revealing the answer!

Final Answer:

The slope of the given line is $m = 2$. Using point-slope form with (1, -3):

$y + 3 = 2(x - 1)$

Simplifying: $y = 2x - 5$

This is the equation of the line parallel to $2x - y = 16$ passing through (1, -3).

Q1(b). Given $f(x) = \frac{2 - x}{x}$, find and simplify the difference quotient $\frac{f(x + h) - f(x)}{h}$, $h \neq 0$.

Background

Topic: Difference Quotient

This question tests your ability to compute and simplify the difference quotient, which is foundational for understanding derivatives in calculus.

Key Terms and Formulas

• Difference Quotient: $\frac{f(x + h) - f(x)}{h}$

Step-by-Step Guidance

1. Find $f(x + h)$ by substituting $x + h$ into the function: $f(x + h) = \frac{2 - (x + h)}{x + h}$.

2. Write the difference $f(x + h) - f(x)$ using the expressions for $f(x + h)$ and $f(x)$.

3. Combine the two fractions over a common denominator.

4. Simplify the numerator as much as possible, but stop before dividing by $h$ and fully simplifying.

Try solving on your own before revealing the answer!

Final Answer:

$f(x + h) = \frac{2 - x - h}{x + h}$

So, $\frac{f(x + h) - f(x)}{h} = \frac{\frac{2 - x - h}{x + h} - \frac{2 - x}{x}}{h}$

Combine over a common denominator and simplify:

$= \frac{(2 - x - h)x - (2 - x)(x + h)}{h(x + h)x}$

$= \frac{(2x - x^2 - hx) - [2x + 2h - x^2 - xh]}{h(x + h)x}$

$= \frac{-h x - 2h}{h(x + h)x}$

$= \frac{-x - 2}{(x + h)x}$

Q2(a). Determine the equation and slope of a horizontal line through the point (4, -2).

Background

Topic: Equations of Horizontal Lines

This question tests your understanding of the properties of horizontal lines and how to write their equations.

Key Terms and Formulas

• Horizontal line: Has the form $y = c$ where $c$ is a constant.

• Slope of a horizontal line: $m = 0$

Step-by-Step Guidance

1. Recall that a horizontal line has a constant $y$ value for all $x$.

2. Use the $y$-coordinate of the given point to write the equation.

3. State the slope of any horizontal line.

Try solving on your own before revealing the answer!

Final Answer:

The equation is $y = -2$ and the slope is $0$.

Q2(b). If the discriminant of a quadratic equation has a negative value, the equation has how many real solutions?

Background

Topic: Quadratic Equations and the Discriminant

This question tests your understanding of the discriminant and its implications for the nature of the solutions of a quadratic equation.

Key Terms and Formulas

• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Discriminant: $D = b^2 - 4ac$

Step-by-Step Guidance

1. Recall the quadratic formula and identify the discriminant.

2. Consider what happens when $D < 0$ (the value under the square root is negative).

3. Think about whether real solutions exist when the square root is of a negative number.

Try solving on your own before revealing the answer!

Final Answer:

There are no real solutions (0 real solutions) when the discriminant is negative.

Q2(c). State the domain of the function $g(x) = \frac{1}{x - 1} + 1$, using interval notation.

Background

Topic: Domain of Rational Functions

This question tests your ability to determine the domain of a function, especially when the function involves division by a variable expression.

Key Terms and Formulas

• Domain: The set of all real numbers for which the function is defined.

• Undefined points: Occur when the denominator is zero.

Step-by-Step Guidance

1. Set the denominator $x - 1$ equal to zero and solve for $x$.

2. Exclude this value from the set of all real numbers.

3. Write the domain in interval notation, but stop before writing the final interval.

Try solving on your own before revealing the answer!

Final Answer:

The domain is all real numbers except $x = 1$, or $(-\infty, 1) \cup (1, \infty)$.

Q3(a). Solve for $x$: $x + 2 = 10$

Background

Topic: Solving Linear Equations

This question tests your ability to solve simple linear equations for a variable.

Key Terms and Formulas

• Linear equation: An equation of the form $ax + b = c$.

Step-by-Step Guidance

1. Isolate $x$ by subtracting $2$ from both sides of the equation.

2. Write the resulting equation, but stop before stating the value of $x$.

Try solving on your own before revealing the answer!

Final Answer:

$x = 8$

Q3(b). Solve for $x$: $\frac{1}{2}x - 1 = \frac{1}{3}x + 2$

Background

Topic: Solving Linear Equations with Fractions

This question tests your ability to solve linear equations that involve fractions.

Key Terms and Formulas

• Linear equation: An equation of the form $ax + b = cx + d$.

Step-by-Step Guidance

1. Subtract $\frac{1}{3}x$ from both sides to get all $x$ terms on one side.

2. Add $1$ to both sides to get all constants on the other side.

3. Combine like terms and stop before dividing to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer:

$x = 9$

Q3(c). Solve for $x$: $4^{1 - 3x} = 16$

Background

Topic: Exponential Equations

This question tests your ability to solve equations where the variable is in the exponent.

Key Terms and Formulas

• Exponential equation: An equation where the variable appears in the exponent.

• Properties of exponents: $a^{m} = a^{n} \implies m = n$ if $a \neq 0$.

Step-by-Step Guidance

1. Express $16$as a power of $4$.

2. Set the exponents equal to each other since the bases are the same.

3. Solve the resulting linear equation for $x$, but stop before stating the value.

Try solving on your own before revealing the answer!

Final Answer:

$x = -\frac{1}{3}$

Q3(d). Solve for $x$: $A - Bx = 9$

Background

Topic: Solving Linear Equations with Parameters

This question tests your ability to solve for a variable in terms of other parameters.

Key Terms and Formulas

• Linear equation: An equation of the form $ax + b = c$.

Step-by-Step Guidance

1. Subtract $A$ from both sides to isolate the $Bx$ term.

2. Divide both sides by $-B$ to solve for $x$, but stop before simplifying the expression.

Try solving on your own before revealing the answer!

Final Answer:

$x = \frac{A - 9}{B}$

Q4. From the graph of a function $h$ to the right (the entire function is pictured), determine the following (estimating to the nearest tenth when necessary):

• (a) the domain of $h$

• (b) the range of $h$

• (c) the $y$-intercept

• (d) the values of $x$ for which $f(x) = 0$

• (e) the interval(s) on which $h$ is increasing

Background

Topic: Reading Graphs of Functions

This question tests your ability to interpret a function's graph to determine its domain, range, intercepts, zeros, and intervals of increase.

Key Terms and Formulas

• Domain: All $x$-values for which the function is defined.

• Range: All $y$-values the function attains.

• $y$-intercept: The point where the graph crosses the $y$-axis ($x = 0$).

• Zeros: $x$-values where $f(x) = 0$ (where the graph crosses the $x$-axis).

• Increasing interval: Where the graph rises as $x$ increases.

Step-by-Step Guidance

1. Examine the leftmost and rightmost points of the graph to estimate the domain (the $x$-values covered).

2. Look at the lowest and highest points on the graph to estimate the range (the $y$-values covered).

3. Find the point where the graph crosses the $y$-axis to determine the $y$-intercept.

4. Identify the $x$-values where the graph crosses the $x$-axis for the zeros.

5. Observe the intervals where the graph is moving upward as you move from left to right to find where $h$ is increasing. Stop before listing the exact intervals.

Try solving on your own before revealing the answer!

Final Answer:

• (a) Domain: $[-2, 3]$

• (b) Range: $[-2, 2]$

• (c) $y$-intercept: $(0, 1)$

• (d) $f(x) = 0$ at $x \approx -1.2$ and $x \approx 2.2$

• (e) $h$ is increasing on $[-2, 0]$ and $[1, 2.2]$

Q5. A farmer has 200 feet of fencing to enclose three adjacent corrals. What measurements will produce an enclosed area of 1200 square feet? Set up a mathematical model (equation), stating clearly what each variable represents, then do the work to answer the question.

Background

Topic: Optimization and Mathematical Modeling

This question tests your ability to set up and solve an optimization problem involving area and perimeter constraints.

Key Terms and Formulas

• Area of a rectangle: $A = l \times w$

• Perimeter for three adjacent corrals: $2l + 4w = 200$ (since there are 2 lengths and 4 widths)

Step-by-Step Guidance

1. Let $l$ be the length and $w$ the width of each corral.

2. Write the area equation: $l \times w = 1200$.

3. Write the perimeter equation for three adjacent corrals: $2l + 4w = 200$.

4. Solve one equation for one variable and substitute into the other, but stop before solving for the final values.

Try solving on your own before revealing the answer!

Final Answer:

Let $l$ be the length and $w$ the width. $l \times w = 1200$, $2l + 4w = 200$.

Solving: $l = 1200/w$, $2(1200/w) + 4w = 200$

$2400/w + 4w = 200$

Multiply both sides by $w$ and solve the quadratic: $4w^2 - 200w + 2400 = 0$

Solutions: $w = 12$ ft, $l = 100$ ft; or $w = 50$ ft, $l = 24$ ft (only the one that fits the perimeter constraint is valid: $w = 12$ ft, $l = 100$ ft).

Q6(a). Given $m(x) = \frac{2x^2 + 3x + 9}{x - 2}$, find $m(3)$.

Background

Topic: Evaluating Rational Functions

This question tests your ability to substitute a value into a rational function and simplify.

Key Terms and Formulas

• Substitution: Replace $x$ with the given value in the function.

Step-by-Step Guidance

1. Substitute $x = 3$ into the numerator and denominator.

2. Compute the numerator and denominator separately, but stop before dividing.

Try solving on your own before revealing the answer!

Final Answer:

$m(3) = \frac{2(3)^2 + 3(3) + 9}{3 - 2} = \frac{18 + 9 + 9}{1} = 36$

Q6(b). Simplify $\frac{3x^2 - 5x + 2}{x - 3}$.

Background

Topic: Simplifying Rational Expressions

This question tests your ability to factor and simplify rational expressions.

Key Terms and Formulas

• Factoring: Expressing a polynomial as a product of its factors.

Step-by-Step Guidance

1. Factor the numerator $3x^2 - 5x + 2$.

2. Check if any factors cancel with the denominator $x - 3$.

3. Write the simplified expression, but stop before canceling or simplifying completely.

Try solving on your own before revealing the answer!

Final Answer:

$3x^2 - 5x + 2 = (3x - 2)(x - 1)$, so $\frac{3x^2 - 5x + 2}{x - 3}$ does not simplify further since $x - 3$ is not a factor of the numerator.

Q6(c). Simplify, writing the result with only positive exponents: $\left(\frac{2x^5}{3x^2}\right)^{-2}$.

Background

Topic: Exponents and Negative Exponents

This question tests your ability to simplify expressions with negative exponents and write the result with only positive exponents.

Key Terms and Formulas

• Negative exponent rule: $a^{-n} = \frac{1}{a^n}$

• Exponent rules: $\frac{a^m}{a^n} = a^{m-n}$

Step-by-Step Guidance

1. Simplify the inside of the parentheses first: $\frac{2x^5}{3x^2}$.

2. Apply the negative exponent to the entire fraction.

3. Rewrite the expression with only positive exponents, but stop before writing the final simplified form.

Try solving on your own before revealing the answer!

Final Answer:

$\left(\frac{2x^5}{3x^2}\right)^{-2} = \left(\frac{2}{3}x^{3}\right)^{-2} = \frac{3^2}{2^2}x^{-6} = \frac{9}{4x^6}$

Q6(d). Simplify $\sqrt{36x^4 + 4}$.

Background

Topic: Simplifying Radical Expressions

This question tests your ability to simplify square roots, especially when the radicand is a sum.

Key Terms and Formulas

• Square root: $\sqrt{a^2} = |a|$

Step-by-Step Guidance

1. Factor out any perfect squares from the radicand if possible.

2. Check if the expression can be written as a sum of squares or factored further, but stop before fully simplifying.

Try solving on your own before revealing the answer!

Final Answer:

$\sqrt{36x^4 + 4} = \sqrt{4(9x^4 + 1)} = 2\sqrt{9x^4 + 1}$

Q6(e). Simplify without using a calculator: $0.05 \times 250$.

Background

Topic: Basic Arithmetic

This question tests your ability to multiply decimals and whole numbers without a calculator.

Key Terms and Formulas

• Decimal multiplication: $a \times b$

Step-by-Step Guidance

1. Rewrite $0.05$ as a fraction.

2. Multiply the fraction by $250$ and simplify, but stop before stating the final value.

Try solving on your own before revealing the answer!

Final Answer:

$0.05 \times 250 = \frac{5}{100} \times 250 = \frac{1}{20} \times 250 = 12.5$

Q7(a). A model rocket is fired upward from ground level at an initial velocity of 60 feet per second. (Use the position equation $s = s_0 + v_0 t - 16t^2$) Determine how long it will take for the rocket to come back down to the ground.

Background

Topic: Quadratic Equations in Physics

This question tests your ability to use the position equation for vertical motion to solve for time when the object returns to the ground.

Key Terms and Formulas

• Position equation: $s = s_0 + v_0 t - 16t^2$

• Initial position: $s_0 = 0$ (ground level)

• Initial velocity: $v_0 = 60$ ft/s

Step-by-Step Guidance

1. Set $s = 0$ (when the rocket returns to the ground).

2. Write the equation: $0 = 0 + 60t - 16t^2$.

3. Rearrange to standard quadratic form and factor or use the quadratic formula, but stop before solving for $t$.

Try solving on your own before revealing the answer!

Final Answer:

$0 = 60t - 16t^2 \implies 16t^2 - 60t = 0 \implies t(16t - 60) = 0$

So $t = 0$ (launch) or $t = 3.75$ seconds (when it returns to the ground).

Q7(b). How high above ground is the rocket after 2 seconds?

Background

Topic: Evaluating Quadratic Functions in Physics

This question tests your ability to substitute a value into the position equation to find the height at a given time.

Key Terms and Formulas

• Position equation: $s = s_0 + v_0 t - 16t^2$

Step-by-Step Guidance

1. Substitute $t = 2$ into the position equation with $s_0 = 0$ and $v_0 = 60$.

2. Calculate each term separately, but stop before adding them together for the final height.

Try solving on your own before revealing the answer!

Final Answer:

$s = 0 + 60(2) - 16(2)^2 = 120 - 64 = 56$ feet