$f(x) = |ax + b| + c$ Graph: V-shape. The vertex is at $(-b/a, c)$.

Square Root Functions

$f(x) = \sqrt{ax + b}$ Graph: Half-parabola (starts at a point, curves right). Domain: $ax + b \geq 0$

Rational Functions

$f(x) = \frac{a}{x - h} + k$ Graph: Hyperbola with asymptotes at $x = h$ and $y = k$.

Solving Systems with Nonlinear Equations

Systems involving one linear and one nonlinear equation:

Linear-Quadratic System

$y = x + 2 \quad \text{and} \quad y = x^2$

Substitute: $x^2 = x + 2$ $x^2 - x - 2 = 0$ $(x-2)(x+1) = 0$ $x = 2 \text{ or } x = -1$

Solutions: $(2, 4)$ and $(-1, 1)$

Number of solutions:

• The line can intersect the parabola at 0, 1, or 2 points
• 0 points: no solution
• 1 point: tangent line
• 2 points: two solutions

Radical Equations

Solving Equations with Square Roots

Strategy: Isolate the radical, then square both sides.

$\sqrt{2x + 3} = 5$ $2x + 3 = 25$ $2x = 22$ $x = 11$

Always check! Squaring can introduce extraneous solutions.

Example with Extraneous Solution

$\sqrt{x + 5} = x - 1$ $x + 5 = (x-1)^2 = x^2 - 2x + 1$ $0 = x^2 - 3x - 4 = (x-4)(x+1)$ $x = 4 \text{ or } x = -1$

Check $x = 4$: $\sqrt{9} = 3$ and $4 - 1 = 3$ ✓ Check $x = -1$: $\sqrt{4} = 2$ and $-1 - 1 = -2$ ✗ (extraneous!)

Absolute Value Equations

$|ax + b| = c$

If $c \geq 0$: Two equations → $ax + b = c$ or $ax + b = -c$ If $c < 0$: No solution (absolute value can't be negative)

Example: $|2x - 3| = 7$ $2x - 3 = 7 \implies x = 5$ $2x - 3 = -7 \implies x = -2$

Function Composition and Evaluation

For complex function problems:

1. Read carefully — what specific value or expression are they asking for?
2. Substitute step by step
3. Simplify completely

SAT Question Types

Type 1: Solve a Radical Equation

Isolate the radical, square both sides, check for extraneous solutions.

Type 2: Linear-Quadratic System

Substitute and solve the resulting quadratic.

Type 3: Number of Intersections

Use the discriminant of the resulting quadratic to determine 0, 1, or 2 intersections.

Type 4: Absolute Value Equations

Split into two cases.

Common SAT Mistakes

1. Not checking for extraneous solutions in radical equations
2. Forgetting there are two cases for absolute value
3. Errors when squaring both sides — expand $(x-1)^2$ carefully!
4. Assuming a nonlinear system always has 2 solutions — it could have 0 or 1
5. Domain errors — $\sqrt{x}$ requires $x \geq 0$

Case 2: $x - 4 = -6 \implies x = -2$

Check: $|10 - 4| = |6| = 6$ ✓ and $|-2 - 4| = |-6| = 6$ ✓

Answer: $x = 10$ or $x = -2$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 3: Check: $\sqrt{3(5) + 1} = \sqrt{16} = 4$ ✓

Answer: $x = 5$

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$, the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

• $\Delta > 0$ → 2 intersections
• $\Delta = 0$ → 1 intersection (tangent)
• $\Delta < 0$ → 0 intersections

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$, the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

• $\Delta > 0$ → 2 intersections
• $\Delta = 0$ → 1 intersection (tangent)
• $\Delta < 0$ → 0 intersections

Step 2: Use the discriminant: $b^2 - 4ac = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$

Since $\Delta = 12 > 0$, the quadratic has two real solutions, meaning the line intersects the parabola at two points.

Answer: 2 solutions

Discriminant shortcut:

• $\Delta > 0$ → 2 intersections
• $\Delta = 0$ → 1 intersection (tangent)
• $\Delta < 0$ → 0 intersections

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$: $\sqrt{-3 + 7} = \sqrt{4} = 2$ and $-3 + 1 = -2$. Is $2 = -2$? No! ✗ Extraneous!

$x = 2$: $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 + 1 = 3$. Is $3 = 3$? Yes! ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$: $\sqrt{-3 + 7} = \sqrt{4} = 2$ and $-3 + 1 = -2$. Is $2 = -2$? No! ✗ Extraneous!

$x = 2$: $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 + 1 = 3$. Is $3 = 3$? Yes! ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 2: Rearrange: $0 = x^2 + 2x + 1 - x - 7 = x^2 + x - 6$

Step 3: Factor: $(x + 3)(x - 2) = 0$ $x = -3 \text{ or } x = 2$

Step 4: CHECK both (squaring can create extraneous solutions):

$x = -3$: $\sqrt{-3 + 7} = \sqrt{4} = 2$ and $-3 + 1 = -2$. Is $2 = -2$? No! ✗ Extraneous!

$x = 2$: $\sqrt{2 + 7} = \sqrt{9} = 3$ and $2 + 1 = 3$. Is $3 = 3$? Yes! ✓

Answer: $x = 2$ only

Key lesson: Always check solutions in radical equations!

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$ $9 - 8 + 4k = 0$ $1 + 4k = 0$ $k = -\frac{1}{4}$

Step 3: Verify: With $k = -\frac{1}{4}$: $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$ $9 - 8 + 4k = 0$ $1 + 4k = 0$ $k = -\frac{1}{4}$

Step 3: Verify: With $k = -\frac{1}{4}$: $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$

Step 2: For exactly one intersection, the discriminant must equal zero: $b^2 - 4ac = 0$ $(-3)^2 - 4(1)(2 - k) = 0$ $9 - 8 + 4k = 0$ $1 + 4k = 0$ $k = -\frac{1}{4}$

Step 3: Verify: With $k = -\frac{1}{4}$: $x^2 - 3x + \frac{9}{4} = 0 \implies (x - \frac{3}{2})^2 = 0 \implies x = \frac{3}{2}$

One solution ✓

Answer: $k = -\frac{1}{4}$