Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  2. SAT Math
  3. Flashcards

SAT Math Flashcards: Graphing Functions

Study Graphing Functions in SAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Graphing Functions, giving you a quick way to review the definitions, rules, and examples that matter most for SAT Math.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

SAT Math Flashcards: Graphing Functions

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

Identify the range of the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣.

Tap or drag to reveal answer

ANSWER

y≥0y \geq 0y≥0. Absolute value function always produces non-negative outputs.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: Identify the range of the function f(x)=∣x∣f(x) = |x|f(x)=∣x∣.

Answer: y≥0y \geq 0y≥0. Absolute value function always produces non-negative outputs.

Flashcard 2: What is the general form of the exponential function?

Answer: y=abxy = ab^xy=abx. Base bbb raised to variable power xxx with coefficient aaa.

Flashcard 3: What is the shape of the graph of y=1xy = \frac{1}{x}y=x1​?

Answer: Hyperbola. Rational function with two branches in opposite quadrants.

Flashcard 4: What is the general form of a linear function?

Answer: y=mx+by = mx + by=mx+b. Slope-intercept form where mmm is slope and bbb is y-intercept.

Flashcard 5: State the equation of a horizontal line.

Answer: y=cy = cy=c. Horizontal lines have zero slope and constant y-value.

Flashcard 6: What is the range of the function y=ln⁡(x)y = \ln(x)y=ln(x)?

Answer: All real numbers. Natural logarithm outputs all real values for positive inputs.

Flashcard 7: What is the result of f(x)=x3f(x) = x^3f(x)=x3 being reflected over the y-axis?

Answer: f(x)=−x3f(x) = -x^3f(x)=−x3. Reflection over y-axis changes xxx to −x-x−x in odd functions.

Flashcard 8: Find the domain of the function f(x)=x−4f(x) = \sqrt{x - 4}f(x)=x−4​.

Answer: x≥4x \geq 4x≥4. Square root requires non-negative argument: x−4≥0x - 4 \geq 0x−4≥0.

Flashcard 9: What is the slope of the line y=−5x+2y = -5x + 2y=−5x+2?

Answer: m=−5m = -5m=−5. Coefficient of xxx in linear form y=mx+by = mx + by=mx+b.

Flashcard 10: Find the y-intercept of the function f(x)=4x2−3x+5f(x) = 4x^2 - 3x + 5f(x)=4x2−3x+5.

Answer: y=5y = 5y=5. Y-intercept is the constant term when x=0x = 0x=0.

Flashcard 11: State the domain for f(x)=1x2−x−6f(x) = \frac{1}{x^2 - x - 6}f(x)=x2−x−61​.

Answer: x≠3,x≠−2x \neq 3, x \neq -2x=3,x=−2. Factor: (x−3)(x+2)≠0(x-3)(x+2) \neq 0(x−3)(x+2)=0, so exclude both roots.

Flashcard 12: Find the x-intercept of the line: y=3x+6y = 3x + 6y=3x+6.

Answer: x=−2x = -2x=−2. Set y=0y = 0y=0 and solve: 0=3x+60 = 3x + 60=3x+6, so x=−2x = -2x=−2.

Flashcard 13: Find the vertex of y=2(x−3)2+4y = 2(x - 3)^2 + 4y=2(x−3)2+4.

Answer: (3,4)(3, 4)(3,4). Vertex form directly shows vertex coordinates (h,k)(h, k)(h,k).

Flashcard 14: Identify the vertex form of a quadratic function.

Answer: y=a(x−h)2+ky = a(x - h)^2 + ky=a(x−h)2+k. Shows vertex at (h,k)(h, k)(h,k) with horizontal shifts and vertical shifts.

Flashcard 15: Identify the equation for a circle centered at the origin.

Answer: x2+y2=r2x^2 + y^2 = r^2x2+y2=r2. Circle equation with center (0,0)(0,0)(0,0) and radius rrr.

Flashcard 16: State the equation for the asymptote of y=2x−3y = 2^x - 3y=2x−3.

Answer: y=−3y = -3y=−3. Exponential functions approach horizontal asymptotes as x→−∞x \to -\inftyx→−∞.

Flashcard 17: Identify the range of the function f(x)=−2x2+4f(x) = -2x^2 + 4f(x)=−2x2+4.

Answer: y≤4y \leq 4y≤4. Parabola opens downward with maximum value y=4y = 4y=4.

Flashcard 18: What is the y-intercept of the line: y=−4x+7y = -4x + 7y=−4x+7?

Answer: y=7y = 7y=7. Y-intercept occurs when x=0x = 0x=0, giving y=7y = 7y=7.

Flashcard 19: Identify the range of y=3x2−5y = 3x^2 - 5y=3x2−5.

Answer: y≥−5y \geq -5y≥−5. Parabola opens upward with minimum value at vertex y=−5y = -5y=−5.

Flashcard 20: What is the slope of the line y=−5x+2y = -5x + 2y=−5x+2?

Answer: m=−5m = -5m=−5. Coefficient of xxx in linear form y=mx+by = mx + by=mx+b.

Flashcard 21: Find the domain of the function f(x)=x−4f(x) = \sqrt{x - 4}f(x)=x−4​.

Answer: x≥4x \geq 4x≥4. Square root requires non-negative argument: x−4≥0x - 4 \geq 0x−4≥0.

Flashcard 22: Identify the axis of symmetry for y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c.

Answer: x=−b2ax = -\frac{b}{2a}x=−2ab​. Vertical line through the vertex of any parabola.

Flashcard 23: Identify the transformation: f(x)=x2f(x) = x^2f(x)=x2 to f(x)=(x+5)2f(x) = (x + 5)^2f(x)=(x+5)2.

Answer: Left 5. (x+5)(x + 5)(x+5) represents horizontal shift left by 5 units.

Flashcard 24: What is the effect of f(x)=x2f(x) = x^2f(x)=x2 becoming f(x)=−x2f(x) = -x^2f(x)=−x2?

Answer: Reflection over x-axis. Negative coefficient flips parabola upside down.

Flashcard 25: What is the slope of a vertical line?

Answer: Undefined. Vertical lines have infinite slope, expressed as undefined.

Flashcard 26: Identify the transformation: f(x)=(x−4)2+2f(x) = (x - 4)^2 + 2f(x)=(x−4)2+2.

Answer: Right 4, Up 2. (x−4)(x - 4)(x−4) shifts right 4, +2+2+2 shifts up 2.

Flashcard 27: State the transformation: f(x)=x2f(x) = x^2f(x)=x2 to f(x)=3x2f(x) = 3x^2f(x)=3x2.

Answer: Vertical stretch by 3. Coefficient greater than 1 stretches graph vertically.

Flashcard 28: What is the general form of a linear equation?

Answer: y=mx+by = mx + by=mx+b. Standard linear form with slope mmm and y-intercept bbb.

Flashcard 29: State the formula for the slope of a line through (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​).

Answer: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. Rise over run formula between two points.

Flashcard 30: What is the equation of a parabola opening upwards with vertex (h,k)(h, k)(h,k)?

Answer: y=a(x−h)2+ky = a(x - h)^2 + ky=a(x−h)2+k. Vertex form with a>0a > 0a>0 opens upward from vertex (h,k)(h, k)(h,k).