🎯⭐ INTERACTIVE LESSON

Intermediate Algebra and Coordinate Geometry

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Intermediate Algebra and Coordinate Geometry - Complete Interactive Lesson

Part 1: Exponents, Radicals & Algebraic Expressions

📐 Intermediate Algebra & Coordinate Geometry

Part 1 of 5 — Exponents, Radicals & Algebraic Expressions

Topics in This Part

Section
The Exponent Rules
Negative & Zero Exponents
Simplifying Radicals
Rationalizing the Denominator

🔑 Key Concept: The ACT Intermediate Algebra & Coordinate Geometry reporting category is worth roughly 18 of the 60 math questions — the single biggest slice. It rewards students who can manipulate exponents and radicals quickly and read the $xy$ -plane fluently. Part 1 locks down the algebra of powers and roots.

The Exponent Rules

Every exponent question on the ACT reduces to these five rules:

Rule	Statement	Example
Product	$x^a \cdot x^b = x^{a+b}$

Concept Check 🎯

Negative & Zero Exponents

Two special cases trip up a lot of students:

$x^0 = 1 \quad (x \ne 0) \qquad\qquad x^{-n} = \frac{1}{x^n}$

Evaluate the Powers 🧮

Enter each value as an integer or fraction (e.g. 1/8).

1) $4^0 + 3^0 = \,?$ 2) $2^{-4} = \,?$

Simplifying Radicals

To simplify a square root, pull out the largest perfect-square factor:

$\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}$

Concept Check 🎯

Part 2: Quadratics: Factoring, the Formula & Systems

📐 Intermediate Algebra & Coordinate Geometry

Part 2 of 5 — Quadratics: Factoring, the Formula & Systems

🔑 The Idea: A quadratic $ax^2 + bx + c = 0$ can be solved three ways — factoring, the quadratic formula, and (sometimes) the square-root method. The ACT expects you to pick the fastest one and to know the link between roots and factors.

Factoring & the Zero-Product Property

Part 3: Inequalities, Functions & Absolute Value

📐 Intermediate Algebra & Coordinate Geometry

Part 3 of 5 — Inequalities, Functions & Absolute Value

🔑 Why it matters: ACT questions love function notation $f(x)$ and inequalities. The one rule you can't forget: multiplying or dividing an inequality by a negative number flips the inequality sign.

Solving Inequalities

Solve an inequality exactly like an equation — with one extra rule.

⚠️ Flip the sign whenever you multiply or divide both sides by a negative number.

Worked example: solve $-3x + 5 < 14$ .

Part 4: The Coordinate Plane: Slope, Distance & Lines

📐 Intermediate Algebra & Coordinate Geometry

Part 4 of 5 — The Coordinate Plane: Slope, Distance & Lines

🔑 Big Payoff: "Coordinate geometry" is half the name of this reporting category. Master the three core formulas — slope, distance, and midpoint — and the equation of a line, and you own a huge block of ACT points.

The Three Core Formulas

For two points $(x_1, y_1)$ and :

Part 5: Circles, Mixed Practice & Mastery Check

📐 Intermediate Algebra & Coordinate Geometry

Part 5 of 5 — Circles, Mixed Practice & Mastery Check

You can now handle exponents, radicals, quadratics, inequalities, functions, and lines in the plane. Part 5 adds the circle and then mixes everything for a final check.

Circles in the Coordinate Plane

The standard form of a circle with center $(h, k)$ and radius $r$ is:

$(x - h)^2 + (y - k)^2 = r^2.$

⚠️ Trap: you add exponents when you multiply like bases, but multiply exponents when you raise a power to a power. Mixing these two up is the most common exponent error on the test.

Worked example: simplify $\dfrac{(2x^2)^3}{4x^4}$ .

$\frac{(2x^2)^3}{4x^4} = \frac{8x^6}{4x^4} = 2x^{6-4} = 2x^2$

A negative exponent is a signal to flip the base across the fraction bar — it does not make the value negative.

Expression	Rewrite	Value
$5^0$	—	$1$
$2^{-3}$	$\dfrac{1}{2^3}$	$\dfrac{1}{8}$
$\dfrac{1}{x^{-2}}$	$x^2$
$\left(\dfrac{3}{4}\right)^{-1}$	$\dfrac{4}{3}$

💡 Fraction shortcut: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}$ — a negative exponent just turns a fraction upside down.

\left(\dfrac{2}{5}\right)^{-2} = \,?

Worked example: $\sqrt{72}$ . Since $72 = 36 \cdot 2$ and $36$ is a perfect square,

$\sqrt{72} = \sqrt{36}\cdot\sqrt{2} = 6\sqrt{2}.$

Rationalizing the Denominator

The ACT prefers answer choices with no radical in the denominator. Multiply top and bottom by the radical:

$\frac{3}{\sqrt{2}} = \frac{3}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.$

🔑 Fractional exponents are radicals in disguise: $x^{1/2} = \sqrt{x}$ and $x^{m/n} = \sqrt[n]{x^{\,m}}$ . So $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$ .

When $ax^2 + bx + c$ factors with $a = 1$ , find two numbers that multiply to $c$ and add to $b$ .

Worked example: solve $x^2 + 2x - 15 = 0$ .

We need two numbers with product $-15$ and sum $+2$ : those are $+5$ and $-3$ .

$x^2 + 2x - 15 = (x+5)(x-3) = 0$

By the zero-product property, set each factor to $0$ :

$x + 5 = 0 \;\Rightarrow\; x = -5 \qquad x - 3 = 0 \;\Rightarrow\; x = 3.$

💡 Roots ⇄ factors: if $x = r$ is a root, then $(x - r)$ is a factor. So a quadratic with roots $4$ and $-1$ is $(x-4)(x+1) = x^2 - 3x - 4$ .

Concept Check 🎯

The Quadratic Formula

When a quadratic won't factor nicely, use:

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

The quantity under the radical, $b^2 - 4ac$ , is the discriminant. It tells you how many real solutions exist:

Discriminant $b^2 - 4ac$	Real solutions
Positive	2 distinct real roots
Zero	1 repeated real root
Negative	0 real roots (two complex)

Worked example: solve $x^2 + 4x + 1 = 0$ ( $a=1, b=4, c=1$ ).

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3}.$

⚠️ Watch the signs when computing $b^2 - 4ac$ : if $c$ is negative, $-4ac$ becomes positive, which enlarges the discriminant.

Read the Discriminant 🔽

For each quadratic, decide how many real solutions it has.

Use the Formula 🧮

1) For $2x^2 - 3x - 5 = 0$ , compute the discriminant $b^2 - 4ac = \,?$ 2) Solve $2x^2 - 3x - 5 = 0$ and enter the larger root. (Hint: it factors.)

Systems of Equations

A system is two equations solved together. Use substitution or elimination.

Worked example (elimination): solve $\begin{cases} 3x + 2y = 12 \\ x - 2y = 4 \end{cases}$

Add the equations to eliminate $y$ : $\;4x = 16 \Rightarrow x = 4$ .

Substitute back: $4 - 2y = 4 \Rightarrow y = 0$ . Solution: $(4, 0)$ .

💡 Graphical meaning: the solution of a linear system is the point where the two lines cross. Parallel lines (same slope, different intercept) never cross → no solution.

$-3x + 5 < 14 \;\Rightarrow\; -3x < 9 \;\Rightarrow\; x > -3$

The division by $-3$ flipped $<$ into $>$ .

Compound Inequalities

A statement like $-2 \le 2x - 4 < 6$ means both parts hold. Operate on all three pieces at once:

$-2 \le 2x - 4 < 6 \;\Rightarrow\; 2 \le 2x < 10 \;\Rightarrow\; 1 \le x < 5.$

Concept Check 🎯

Function Notation

$f(x)$ is a machine: whatever goes in the parentheses replaces every $x$ .

Worked example: if $f(x) = x^2 - 3x + 2$ , find $f(-2)$ .

$f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12.$

Composition $f(g(x))$ means "do $g$ first, then feed the result into $f$ ."

Worked example: if $f(x) = 2x + 1$ and $g(x) = x^2$ , find .

$g(3) = 3^2 = 9, \qquad f(9) = 2(9) + 1 = 19.$

🔑 Work inside-out: evaluate the innermost parentheses first, then substitute outward.

Evaluate the Functions 🧮

Let $f(x) = x^2 - 5$ and $g(x) = 3x + 1$ .

1) $f(4) = \,?$ 2) $g(-2) = \,?$ 3) $f(g(0)) = \,?$

Absolute-Value Equations

$|x|$ measures distance from zero, so it's never negative. To solve $|expression| = k$ (with $k \ge 0$ ), split into two cases:

$|x - 3| = 7 \;\Rightarrow\; x - 3 = 7 \;\text{ or }\; x - 3 = -7$

⚠️ If the absolute value equals a negative number, e.g. $|x| = -5$ , there is no solution — distance can't be negative.

💡 On a number line, $|x - 3| = 7$ asks "which points are exactly $7$ away from $3$ ?" — namely $3 + 7 = 10$ and .

Absolute Value & Functions 🔽

Formula	Equation
Slope	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
Distance	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Midpoint	$\left(\dfrac{x_1 + x_2}{2},\; \dfrac{y_1 + y_2}{2}\right)$

💡 Slope is a difference, midpoint is an average, and distance is the Pythagorean theorem in disguise.

Worked example: for $A(1, 2)$ and $B(4, 6)$ :

$m = \frac{6 - 2}{4 - 1} = \frac{4}{3}, \qquad d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

$\text{midpoint} = \left(\frac{1+4}{2}, \frac{2+6}{2}\right) = \left(\frac{5}{2}, 4\right).$

Apply the Formulas 🧮

Use the points $P(-2, 1)$ and $Q(4, 9)$ .

1) Slope of $\overline{PQ} = \,?$ 2) Distance $PQ = \,?$ (it's a whole number) 3) Midpoint $x$ -coordinate $= \,?$

Equations of Lines

Slope-intercept form is the ACT workhorse:

$y = mx + b \qquad m = \text{slope}, \; b = y\text{-intercept}.$

Point-slope form is handy when you know a point and a slope:

$y - y_1 = m(x - x_1).$

Parallel & Perpendicular

Relationship	Slopes
Parallel	equal: $m_1 = m_2$
Perpendicular	negative reciprocals: $m_{1} \cdot m_{2} =$

🔑 To find a perpendicular slope, flip and negate. The perpendicular to a line with slope $\dfrac{2}{3}$ has slope $-\dfrac{3}{2}$ .

Concept Check 🎯

Slopes & Lines 🔽

Watch the signs — the center coordinates are the opposite of what's inside each parenthesis.

Equation	Center	Radius
$(x-3)^2 + (y-2)^2 = 25$	$(3, 2)$	$5$
$(x+1)^2 + (y-4)^2 = 9$
$x^2 + y^2 = 16$	$(0, 0)$

⚠️ The right side is $r^2$ , not $r$ . If it equals $25$ , the radius is $\sqrt{25} = 5$ , not $25$ .

Read the Circle 🔽

Quick Reference

Goal	Key move
Multiply like bases	add exponents: $x^a x^b = x^{a+b}$
Power of a power	multiply exponents: $(x^a)^b = x^{ab}$
Solve a quadratic	factor, or $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$
Count real roots	sign of discriminant $b^2 - 4ac$
Solve an inequality	flip the sign when $\times$ or $\div$ by a negative
Slope of a line	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
Perpendicular slope	negative reciprocal (flip & negate)
Circle	$(x-h)^2 + (y-k)^2 = r^2 \Rightarrow$ center , radius

🔑 If you remember only three formulas for coordinate geometry, make them slope, distance ( $=$ Pythagoras), and midpoint ( $=$ average).

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

x = 10 \quad\text{or}\quad x = -4.

m_{1} \cdot m_{2} = - 1