Study Graphs in SAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: What is the slope-intercept form of a linear equation? Answer: y = m x + b y = mx + b y = m x + b . Standard form where m m m is slope and b b b is y-intercept.
Flashcard 2: What is the general form of the equation of a parabola? Answer: y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c . Quadratic with degree 2 polynomial.
Flashcard 3: What is the equation for a line perpendicular to y = − 1 4 x + 2 y = -\frac{1}{4}x + 2 y = − 4 1 x + 2 at ( 0 , 0 ) (0, 0) ( 0 , 0 ) ? Answer: y = 4 x y = 4x y = 4 x . Negative reciprocal slope through given point.
Flashcard 4: Determine the x-intercept of the line y = 2 x − 8 y = 2x - 8 y = 2 x − 8 . Answer:
Set y = 0 y = 0 y = 0 and solve: 0 = 2 x − 8 0 = 2x - 8 0 = 2 x − 8 , so x = 4 x = 4 x = 4 .
Flashcard 5: What is the slope formula for a line passing through points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: m = y 2 − y 1 x 2 − x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 − x 1 y 2 − y 1 . Change in y y y over change in x x x between two points.
Flashcard 6: Identify the slope of the line y − 6 = 1 3 ( x − 3 ) y - 6 = \frac{1}{3}(x - 3) y − 6 = 3 1 ( x − 3 ) . Answer: 1 3 \frac{1}{3} 3 1 . Coefficient of ( x − 3 ) (x - 3) ( x − 3 ) in point-slope form.
Flashcard 7: What is the equation of a vertical line through point ( a , b ) (a, b) ( a , b ) ? Answer: x = a x = a x = a . All points have the same x x x -coordinate.
Flashcard 8: Identify whether the line x = 5 x = 5 x = 5 is vertical or horizontal. Answer: Vertical. Constant x x x -value creates vertical line.
Flashcard 9: Identify the direction of opening for the parabola y = − x 2 + 4 x − 1 y = -x^2 + 4x - 1 y = − x 2 + 4 x − 1 . Answer: Downward. Negative coefficient of x 2 x^2 x 2 opens downward.
Flashcard 10: What type of graph represents the equation y = x 2 y = x^2 y = x 2 ? Answer: Parabola. Quadratic function creates U-shaped curve.
Flashcard 11: What is the equation of a circle with center ( h , k ) (h, k) ( h , k ) and radius r r r ? Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Distance from center to any point on circle.
Flashcard 12: What is the slope of a line with equation y = 0 y = 0 y = 0 ? Answer:
Horizontal line has zero rise over run.
Flashcard 13: Which form of a line's equation uses the formula y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) ? Answer: Point-slope form. Uses a known point ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and slope m m m .
Flashcard 14: Identify the vertex of the parabola y = ( x − 2 ) 2 − 3 y = (x - 2)^2 - 3 y = ( x − 2 ) 2 − 3 . Answer: ( 2 , − 3 ) (2, -3) ( 2 , − 3 ) . Vertex form shows ( h , k ) (h, k) ( h , k ) as the turning point.
Flashcard 15: Identify the y-intercept in the equation y = 3 x + 4 y = 3x + 4 y = 3 x + 4 . Answer:
The constant term when x = 0 x = 0 x = 0 .
Flashcard 16: What are the intercepts in the equation y = 1 2 x − 3 y = \frac{1}{2}x - 3 y = 2 1 x − 3 ? Answer: x-intercept: 6, y-intercept: -3. Set each variable to zero alternately.
Flashcard 17: Determine the x-intercept of the line y = 2 x − 8 y = 2x - 8 y = 2 x − 8 . Answer:
Set y = 0 y = 0 y = 0 and solve: 0 = 2 x − 8 0 = 2x - 8 0 = 2 x − 8 , so x = 4 x = 4 x = 4 .
Flashcard 18: What is the equation of a horizontal line through point ( a , b ) (a, b) ( a , b ) ? Answer: y = b y = b y = b . All points have the same y y y -coordinate.
Flashcard 19: Identify the x-intercept in the equation 2 x − 6 = 0 2x - 6 = 0 2 x − 6 = 0 . Answer:
Solve for x x x when y = 0 y = 0 y = 0 .
Flashcard 20: State the vertex form of a quadratic function. Answer: y = a ( x − h ) 2 + k y = a(x - h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h, k) ( h , k ) and vertical shift k k k .
Flashcard 21: What is the general form of a quadratic function? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . Standard form with highest power of 2.
Flashcard 22: What is the slope formula for a line passing through points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: m = y 2 − y 1 x 2 − x 1 m = \frac{y_2 - y_1}{x_2 - x_1} m = x 2 − x 1 y 2 − y 1 . Change in y y y over change in x x x between two points.
Flashcard 23: Identify the slope in the equation y = − 5 x + 2 y = -5x + 2 y = − 5 x + 2 . Answer: -5. Coefficient of x x x in slope-intercept form.
Flashcard 24: What is the definition of a function in terms of ordered pairs? Answer: Each input has exactly one output. Vertical line test ensures unique outputs.
Flashcard 25: What is the slope of a line with equation y = 0 y = 0 y = 0 ? Answer:
Horizontal line has zero rise over run.
Flashcard 26: Identify the intercepts of the quadratic y = ( x − 3 ) ( x + 2 ) y = (x - 3)(x + 2) y = ( x − 3 ) ( x + 2 ) . Answer: x-intercepts: 3 , − 2 3, -2 3 , − 2 . Set y = 0 y = 0 y = 0 to find where parabola crosses x x x -axis.
Flashcard 27: What is the equation of a circle with center ( h , k ) (h, k) ( h , k ) and radius r r r ? Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Distance from center to any point on circle equals radius.
Flashcard 28: What are the coordinates of the y-intercept in the line 3 x + 4 y = 8 3x + 4y = 8 3 x + 4 y = 8 ? Answer: ( 0 , 2 ) (0, 2) ( 0 , 2 ) . Set x = 0 x = 0 x = 0 and solve for y y y .
Flashcard 29: Identify the slope of the line y − 6 = 1 3 ( x − 3 ) y - 6 = \frac{1}{3}(x - 3) y − 6 = 3 1 ( x − 3 ) . Answer: 1 3 \frac{1}{3} 3 1 . Coefficient of ( x − 3 ) (x - 3) ( x − 3 ) in point-slope form.
Flashcard 30: Identify the vertex of the parabola y = ( x − 2 ) 2 − 3 y = (x - 2)^2 - 3 y = ( x − 2 ) 2 − 3 . Answer: ( 2 , − 3 ) (2, -3) ( 2 , − 3 ) . Vertex form shows ( h , k ) (h, k) ( h , k ) as the turning point.
Flashcard 31: Identify the direction of opening for the parabola y = − x 2 + 4 x − 1 y = -x^2 + 4x - 1 y = − x 2 + 4 x − 1 . Answer: Downward. Negative coefficient of x 2 x^2 x 2 opens downward.
Flashcard 32: Identify the slope of the line: 3 x − 4 y = 12 3x - 4y = 12 3 x − 4 y = 12 . Answer: 3 4 \frac{3}{4} 4 3 . Rearrange to y = 3 4 x − 3 y = \frac{3}{4}x - 3 y = 4 3 x − 3 to identify slope.
Flashcard 33: What is the slope-intercept form of a linear equation? Answer: y = m x + b y = mx + b y = m x + b . Shows slope m m m and y y y -intercept b b b directly.
Flashcard 34: What is the general form of the equation of a circle? Answer: x 2 + y 2 + D x + E y + F = 0 x^2 + y^2 + Dx + Ey + F = 0 x 2 + y 2 + D x + E y + F = 0 . Expanded form of circle equation.
Flashcard 35: Identify the transformation applied: y = ( x − 1 ) 2 + 2 y = (x - 1)^2 + 2 y = ( x − 1 ) 2 + 2 . Answer: Right 1 unit, up 2 units. Vertex form shows horizontal and vertical shifts.
Flashcard 36: What is the standard form of a linear equation? Answer: A x + B y = C Ax + By = C A x + B y = C . Linear equation with integer coefficients and no fractions.
Flashcard 37: What is the equation of a line in slope-intercept form? Answer: y = m x + b y = mx + b y = m x + b . Standard form where m m m is slope and b b b is y-intercept.
Flashcard 38: State the vertex form of a quadratic function. Answer: y = a ( x − h ) 2 + k y = a(x - h)^2 + k y = a ( x − h ) 2 + k . Shows vertex ( h , k ) (h, k) ( h , k ) and vertical shift k k k .
Flashcard 39: Identify the x-intercept in the equation 2 x − 6 = 0 2x - 6 = 0 2 x − 6 = 0 . Answer:
Solve for x x x when y = 0 y = 0 y = 0 .
Flashcard 40: What is the radius of the circle x 2 + y 2 = 16 x^2 + y^2 = 16 x 2 + y 2 = 16 ? Answer: 4 4 4 . Square root of the constant term.
Flashcard 41: What is the axis of symmetry for the parabola y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c ? Answer: x = − b 2 a x = -\frac{b}{2a} x = − 2 a b . Vertical line through the vertex of the parabola.
Flashcard 42: Identify the center of the circle ( x − 3 ) 2 + ( y + 4 ) 2 = 25 (x - 3)^2 + (y + 4)^2 = 25 ( x − 3 ) 2 + ( y + 4 ) 2 = 25 . Answer: ( 3 , − 4 ) (3, -4) ( 3 , − 4 ) . Values of ( h , k ) (h, k) ( h , k ) from standard form.
Flashcard 43: What is the vertex of the parabola y = 3 ( x + 2 ) 2 − 5 y = 3(x + 2)^2 - 5 y = 3 ( x + 2 ) 2 − 5 ? Answer: ( − 2 , − 5 ) (-2, -5) ( − 2 , − 5 ) . Vertex form identifies ( h , k ) (h, k) ( h , k ) directly.
Flashcard 44: What is the midpoint formula for the segment connecting ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: ( x 1 + x 2 2 , y 1 + y 2 2 ) \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) ( 2 x 1 + x 2 , 2 y 1 + y 2 ) . Average of the coordinates of the endpoints.
Flashcard 45: What is the standard form of a linear equation? Answer: A x + B y = C Ax + By = C A x + B y = C . Linear equation with integer coefficients A A A , B B B , and C C C .
Flashcard 46: What is the slope-intercept form of a linear equation? Answer: y = m x + b y = mx + b y = m x + b . Shows slope m m m and y y y -intercept b b b directly.
Flashcard 47: What is the equation for a line perpendicular to y = − 1 4 x + 2 y = -\frac{1}{4}x + 2 y = − 4 1 x + 2 at ( 0 , 0 ) (0, 0) ( 0 , 0 ) ? Answer: y = 4 x y = 4x y = 4 x . Negative reciprocal slope through given point.
Flashcard 48: What is the radius of the circle x 2 + y 2 = 16 x^2 + y^2 = 16 x 2 + y 2 = 16 ? Answer:
Square root of the constant term.
Flashcard 49: What is the standard form of a linear equation? Answer: A x + B y = C Ax + By = C A x + B y = C . Linear equation with integer coefficients and no fractions.
Flashcard 50: What is the equation for a line parallel to y = 3 x − 2 y = 3x - 2 y = 3 x − 2 passing through ( 1 , 4 ) (1, 4) ( 1 , 4 ) ? Answer: y = 3 x + 1 y = 3x + 1 y = 3 x + 1 . Same slope, substitute point to find b b b .
Flashcard 51: Find the y-intercept of the line: 4 x + 5 y = 20 4x + 5y = 20 4 x + 5 y = 20 . Answer:
Set x = 0 x = 0 x = 0 : 5 y = 20 5y = 20 5 y = 20 , so y = 4 y = 4 y = 4 .
Flashcard 52: What is the slope of a line perpendicular to y = 2 x + 3 y = 2x + 3 y = 2 x + 3 ? Answer: − 1 2 -\frac{1}{2} − 2 1 . Negative reciprocal of the original slope.
Flashcard 53: What is the formula for converting a line from standard to slope-intercept form? Answer: Solve A x + B y = C Ax + By = C A x + B y = C for y y y . Isolate y y y by dividing by coefficient of y y y .
Flashcard 54: What is the point of intersection for lines y = 2 x + 1 y = 2x + 1 y = 2 x + 1 and y = − x + 4 y = -x + 4 y = − x + 4 ? Answer: ( 1 , 3 ) (1, 3) ( 1 , 3 ) . Solve the system by setting equations equal.
Flashcard 55: What is the equation of a line in slope-intercept form? Answer: y = m x + b y = mx + b y = m x + b . Standard form where m m m is slope and b b b is y-intercept.
Flashcard 56: Identify the transformation applied: y = ( x − 1 ) 2 + 2 y = (x - 1)^2 + 2 y = ( x − 1 ) 2 + 2 . Answer: Right 1 unit, up 2 units. Vertex form shows horizontal and vertical shifts.
Flashcard 57: Identify the slope of the line: 3 x − 4 y = 12 3x - 4y = 12 3 x − 4 y = 12 . Answer: 3 4 \frac{3}{4} 4 3 . Rearrange to y = 3 4 x − 3 y = \frac{3}{4}x - 3 y = 4 3 x − 3 to identify slope.
Flashcard 58: What is the slope of a line parallel to y = − 4 x + 7 y = -4x + 7 y = − 4 x + 7 ? Answer: -4. Parallel lines have identical slopes.
Flashcard 59: Identify the center of the circle ( x − 3 ) 2 + ( y + 4 ) 2 = 25 (x - 3)^2 + (y + 4)^2 = 25 ( x − 3 ) 2 + ( y + 4 ) 2 = 25 . Answer: ( 3 , − 4 ) (3, -4) ( 3 , − 4 ) . Values of ( h , k ) (h, k) ( h , k ) from standard form.
Flashcard 60: What are the coordinates of the x-intercept in the line 5 x − 2 y = 10 5x - 2y = 10 5 x − 2 y = 10 ? Answer: ( 2 , 0 ) (2, 0) ( 2 , 0 ) . Set y = 0 y = 0 y = 0 and solve for x x x .
Flashcard 61: Find the midpoint of the segment joining ( 1 , 2 ) (1, 2) ( 1 , 2 ) and ( 5 , 6 ) (5, 6) ( 5 , 6 ) . Answer: ( 3 , 4 ) (3, 4) ( 3 , 4 ) . Average the x-coordinates and y-coordinates: 1 + 5 2 , 2 + 6 2 \frac{1+5}{2}, \frac{2+6}{2} 2 1 + 5 , 2 2 + 6 .
Flashcard 62: What is the equation of a circle with center ( h , k ) (h, k) ( h , k ) and radius r r r ? Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Standard circle equation with center ( h , k ) (h, k) ( h , k ) and radius r r r .
Flashcard 63: What is the definition of a function in terms of ordered pairs? Answer: Each input has exactly one output. Vertical line test ensures unique outputs.
Flashcard 64: Identify the y-intercept in the equation y = 3 x + 4 y = 3x + 4 y = 3 x + 4 . Answer:
The constant term when x = 0 x = 0 x = 0 .
Flashcard 65: Identify the vertex of the parabola y = ( x − 3 ) 2 + 2 y = (x - 3)^2 + 2 y = ( x − 3 ) 2 + 2 . Answer: (3, 2). Vertex form ( x − h ) 2 + k (x - h)^2 + k ( x − h ) 2 + k has vertex at ( h , k ) (h, k) ( h , k ) .
Flashcard 66: What is the equation of a horizontal line through point ( a , b ) (a, b) ( a , b ) ? Answer: y = b y = b y = b . All points have the same y y y -coordinate.
Flashcard 67: Identify the radius in the equation ( x − 2 ) 2 + ( y + 1 ) 2 = 49 (x - 2)^2 + (y + 1)^2 = 49 ( x − 2 ) 2 + ( y + 1 ) 2 = 49 . Answer: 7 7 7 . Square root of the constant gives radius.
Flashcard 68: What is the general form of the equation of a circle? Answer: x 2 + y 2 + D x + E y + F = 0 x^2 + y^2 + Dx + Ey + F = 0 x 2 + y 2 + D x + E y + F = 0 . Expanded form of circle equation.
Flashcard 69: State the formula for the distance between two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) . Answer: ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Pythagorean theorem applied to coordinate plane.
Flashcard 70: What is the vertex of the parabola y = 3 ( x + 2 ) 2 − 5 y = 3(x + 2)^2 - 5 y = 3 ( x + 2 ) 2 − 5 ? Answer: ( − 2 , − 5 ) (-2, -5) ( − 2 , − 5 ) . Vertex form identifies ( h , k ) (h, k) ( h , k ) directly.
Flashcard 71: Identify the intercepts of the quadratic y = ( x − 3 ) ( x + 2 ) y = (x - 3)(x + 2) y = ( x − 3 ) ( x + 2 ) . Answer: x-intercepts: 3 , − 2 3, -2 3 , − 2 . Set y = 0 y = 0 y = 0 to find where parabola crosses x x x -axis.
Flashcard 72: State the formula for the distance between two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) . Answer: ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Pythagorean theorem applied to coordinate plane.
Flashcard 73: Identify the slope of the line given by 2 y − 4 x = 8 2y - 4x = 8 2 y − 4 x = 8 . Answer:
Rearrange to y = m x + b y = mx + b y = m x + b form first.
Flashcard 74: Which point lies on the line y = 2 x + 3 y = 2x + 3 y = 2 x + 3 : (1, 5) or (2, 4)? Answer: (1, 5). Substituting: y = 2 ( 1 ) + 3 = 5 y = 2(1) + 3 = 5 y = 2 ( 1 ) + 3 = 5 , so point ( 1 , 5 ) (1, 5) ( 1 , 5 ) satisfies the equation.
Flashcard 75: What does the slope of a line represent in a graph? Answer: The rate of change or steepness of the line. Measures how much y y y changes per unit increase in x x x .
Flashcard 76: What are the intercepts in the equation y = 1 2 x − 3 y = \frac{1}{2}x - 3 y = 2 1 x − 3 ? Answer: x-intercept: 6, y-intercept: -3. Set each variable to zero alternately.
Flashcard 77: What is the general form of the equation of a parabola? Answer: y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c . Quadratic with degree 2 polynomial.
Flashcard 78: What is the equation for a line parallel to y = 3 x − 2 y = 3x - 2 y = 3 x − 2 passing through ( 1 , 4 ) (1, 4) ( 1 , 4 ) ? Answer: y = 3 x + 1 y = 3x + 1 y = 3 x + 1 . Same slope, substitute point to find b b b .
Flashcard 79: What is the slope of a line parallel to y = − 4 x + 7 y = -4x + 7 y = − 4 x + 7 ? Answer: -4. Parallel lines have identical slopes.
Flashcard 80: What does the slope of a line represent in a graph? Answer: The rate of change or steepness of the line. Measures how much y y y changes per unit increase in x x x .
Flashcard 81: What is the equation of a vertical line through point ( a , b ) (a, b) ( a , b ) ? Answer: x = a x = a x = a . All points have the same x x x -coordinate.
Flashcard 82: What is the equation of a circle with center ( h , k ) (h, k) ( h , k ) and radius r r r ? Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Standard circle equation with center ( h , k ) (h, k) ( h , k ) and radius r r r .
Flashcard 83: Identify the radius in the equation ( x − 2 ) 2 + ( y + 1 ) 2 = 49 (x - 2)^2 + (y + 1)^2 = 49 ( x − 2 ) 2 + ( y + 1 ) 2 = 49 . Answer:
Square root of the constant gives radius.
Flashcard 84: Which point lies on the line y = 2 x + 3 y = 2x + 3 y = 2 x + 3 : (1, 5) or (2, 4)? Answer: (1, 5). Substituting: y = 2 ( 1 ) + 3 = 5 y = 2(1) + 3 = 5 y = 2 ( 1 ) + 3 = 5 , so point ( 1 , 5 ) (1, 5) ( 1 , 5 ) satisfies the equation.
Flashcard 85: What is the point of intersection for lines y = 2 x + 1 y = 2x + 1 y = 2 x + 1 and y = − x + 4 y = -x + 4 y = − x + 4 ? Answer: ( 1 , 3 ) (1, 3) ( 1 , 3 ) . Solve the system by setting equations equal.
Flashcard 86: What is the axis of symmetry for the parabola y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c ? Answer: x = − b 2 a x = -\frac{b}{2a} x = − 2 a b . Vertical line through the vertex of the parabola.
Flashcard 87: What is the slope of a line perpendicular to y = 2 x + 3 y = 2x + 3 y = 2 x + 3 ? Answer: − 1 2 -\frac{1}{2} − 2 1 . Negative reciprocal of the original slope.
Flashcard 88: Find the y-intercept of the line: 4 x + 5 y = 20 4x + 5y = 20 4 x + 5 y = 20 . Answer:
Set x = 0 x = 0 x = 0 : 5 y = 20 5y = 20 5 y = 20 , so y = 4 y = 4 y = 4 .
Flashcard 89: Identify the slope of the line given by 2 y − 4 x = 8 2y - 4x = 8 2 y − 4 x = 8 . Answer:
Rearrange to y = m x + b y = mx + b y = m x + b form first.
Flashcard 90: What type of graph represents the equation y = x 2 y = x^2 y = x 2 ? Answer: Parabola. Quadratic function creates U-shaped curve.
Flashcard 91: Which form of a line's equation uses the formula y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 = m ( x − x 1 ) ? Answer: Point-slope form. Uses a known point ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and slope m m m .
Flashcard 92: Identify the slope in the equation y = − 5 x + 2 y = -5x + 2 y = − 5 x + 2 . Answer: -5. Coefficient of x x x in slope-intercept form.
Flashcard 93: What are the coordinates of the x-intercept in the line 5 x − 2 y = 10 5x - 2y = 10 5 x − 2 y = 10 ? Answer: ( 2 , 0 ) (2, 0) ( 2 , 0 ) . Set y = 0 y = 0 y = 0 and solve for x x x .
Flashcard 94: What is the formula for the distance between two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Pythagorean theorem applied to coordinate plane differences.
Flashcard 95: What is the general form of a quadratic function? Answer: a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 . Standard form with highest power of 2.
Flashcard 96: What is the midpoint formula for the segment connecting ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: ( x 1 + x 2 2 , y 1 + y 2 2 ) \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) ( 2 x 1 + x 2 , 2 y 1 + y 2 ) . Average of the coordinates of the endpoints.
Flashcard 97: What is the formula for the distance between two points ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 , y 2 ) ? Answer: Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} Distance = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 . Pythagorean theorem applied to coordinate plane differences.
Flashcard 98: What is the formula for converting a line from standard to slope-intercept form? Answer: Solve A x + B y = C Ax + By = C A x + B y = C for y y y . Isolate y y y by dividing by coefficient of y y y .
Flashcard 99: Find the midpoint of the segment joining ( 1 , 2 ) (1, 2) ( 1 , 2 ) and ( 5 , 6 ) (5, 6) ( 5 , 6 ) . Answer: ( 3 , 4 ) (3, 4) ( 3 , 4 ) . Average the x-coordinates and y-coordinates: 1 + 5 2 , 2 + 6 2 \frac{1+5}{2}, \frac{2+6}{2} 2 1 + 5 , 2 2 + 6 .
Flashcard 100: What is the equation of a circle with center ( h , k ) (h, k) ( h , k ) and radius r r r ? Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Distance from center to any point on circle.