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SAT Math Flashcards: Linear Functions

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

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What this deck covers

This deck focuses on Linear 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: Linear Functions

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QUESTION

What is the point-slope form of a linear equation?

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ANSWER

y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​). Uses a known point (x1,y1)(x_1, y_1)(x1​,y1​) and slope mmm.

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Flashcard 1: What is the point-slope form of a linear equation?

Answer: y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​). Uses a known point (x1,y1)(x_1, y_1)(x1​,y1​) and slope mmm.

Flashcard 2: What is the point-slope form of a linear equation?

Answer: y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​). Uses a known point (x1,y1)(x_1, y_1)(x1​,y1​) and slope mmm.

Flashcard 3: What is the x-intercept of the line 2x+5y=102x + 5y = 102x+5y=10?

Answer:

  1. Set y=0y = 0y=0 to get 2x=102x = 102x=10, so x=5x = 5x=5.

Flashcard 4: State the formula for calculating the slope between two points.

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

Flashcard 5: Calculate the slope of the line y=−23x+4y = -\frac{2}{3}x + 4y=−32​x+4.

Answer: −23-\frac{2}{3}−32​. The coefficient of xxx in slope-intercept form.

Flashcard 6: Identify the slope of the line parallel to y=12x−4y = \frac{1}{2}x - 4y=21​x−4.

Answer: 12\frac{1}{2}21​. Parallel lines have identical slopes.

Flashcard 7: State the standard form of a linear equation.

Answer: Ax+By=CAx + By = CAx+By=C. General form with constants AAA, BBB, and CCC.

Flashcard 8: State the formula for calculating the slope between two points.

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

Flashcard 9: Given point (1,2)(1, 2)(1,2) and slope 4, find the equation in point-slope form.

Answer: y−2=4(x−1)y - 2 = 4(x - 1)y−2=4(x−1). Substitute point coordinates and slope into formula.

Flashcard 10: Convert 3x+4y=123x + 4y = 123x+4y=12 to slope-intercept form.

Answer: y=−34x+3y = -\frac{3}{4}x + 3y=−43​x+3. Isolate yyy by dividing both sides by 4.

Flashcard 11: Write the equation of the line with slope 3 and y-intercept -4.

Answer: y=3x−4y = 3x - 4y=3x−4. Direct substitution into y=mx+by = mx + by=mx+b.

Flashcard 12: What is the x-intercept of the line 2x+5y=102x + 5y = 102x+5y=10?

Answer:

  1. Set y=0y = 0y=0 to get 2x=102x = 102x=10, so x=5x = 5x=5.

Flashcard 13: Which term represents the y-intercept in y=5x+9y = 5x + 9y=5x+9?

Answer:

  1. The constant term is the y-intercept.

Flashcard 14: What is the slope of any vertical line?

Answer: Undefined. Division by zero makes slope undefined.

Flashcard 15: What is the x-intercept of the line y=2x−4y = 2x - 4y=2x−4?

Answer:

  1. Set y=0y = 0y=0 and solve: 0=2x−40 = 2x - 40=2x−4, so x=2x = 2x=2.

Flashcard 16: Find the x-intercept of the line 4x−8y=164x - 8y = 164x−8y=16.

Answer:

  1. Set y=0y = 0y=0 to get 4x=164x = 164x=16, so x=4x = 4x=4.

Flashcard 17: Determine if the lines y=2x+3y = 2x + 3y=2x+3 and y=2x−4y = 2x - 4y=2x−4 are parallel.

Answer: Yes, they are parallel. Both lines have slope 2, so they're parallel.

Flashcard 18: Find the slope of a line through points (2,3)(2, 3)(2,3) and (4,7)(4, 7)(4,7).

Answer:

  1. Use m=7−34−2=42=2m = \frac{7-3}{4-2} = \frac{4}{2} = 2m=4−27−3​=24​=2.

Flashcard 19: What condition must be true for lines to be parallel?

Answer: The slopes are equal. Lines with identical slopes never intersect.

Flashcard 20: Identify the slope in the equation y=−3x+5y = -3x + 5y=−3x+5.

Answer: -3. The coefficient of xxx in slope-intercept form.

Flashcard 21: Identify the y-intercept in the equation y=2x−7y = 2x - 7y=2x−7.

Answer: -7. The constant term in slope-intercept form.

Flashcard 22: Convert y=−2x+5y = -2x + 5y=−2x+5 to standard form.

Answer: 2x+y=52x + y = 52x+y=5. Add 2x2x2x to both sides to get standard form.

Flashcard 23: State the formula for calculating slope given two points.

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

Flashcard 24: State the standard form of a linear equation.

Answer: Ax+By=CAx + By = CAx+By=C. General form with constants AAA, BBB, and CCC.

Flashcard 25: What does the 'm' represent in the equation y=mx+by = mx + by=mx+b?

Answer: Slope of the line. The coefficient of xxx that determines steepness and direction.

Flashcard 26: What is the slope of any horizontal line?

Answer:

  1. No vertical change means zero slope.

Flashcard 27: Find the y-intercept of the line 5x+3y=155x + 3y = 155x+3y=15.

Answer:

  1. Set x=0x = 0x=0 to get 3y=153y = 153y=15, so y=5y = 5y=5.

Flashcard 28: State the formula for finding the slope between two points (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​​. Change in yyy divided by change in xxx.

Flashcard 29: Determine the slope of the line 3x−2y=63x - 2y = 63x−2y=6.

Answer: 32\frac{3}{2}23​. Rearrange to get y=32x−3y = \frac{3}{2}x - 3y=23​x−3.

Flashcard 30: What is the point-slope form of a linear equation?

Answer: y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​). Uses a known point (x1,y1)(x_1, y_1)(x1​,y1​) and slope mmm.