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Quizzes > Mathematics

Graphing Quiz: Distance Formula and y - 3 Practice

Quick, free quiz to test your algebra skills. Instant results and distance formula practice.

Editorial: Review CompletedCreated By: That GuyUpdated Aug 26, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
paper art graphs with coordinate axes data points and relation lines on sky blue background inviting quiz practice.

This graphing quiz helps you practice the distance formula, plot points on the coordinate plane, and read vertical shifts like y - 3. Strengthen related skills with a coordinate plane quiz, dig into curves with graphing functions practice, or refresh line basics in a slope-intercept form quiz.

What is the coordinate of the origin in the Cartesian plane?
(1,0)
(0,1)
(1,1)
(0,0)
The origin is the unique point where the x-axis and y-axis intersect at zero, giving the coordinate (0,0). No other pair of zeros represents the intersection of both axes. The Cartesian coordinate system is defined in reference to this central point. .
In which quadrant is the point (-3, 4) located?
Quadrant IV
Quadrant III
Quadrant II
Quadrant I
Quadrant II contains points with negative x-coordinates and positive y-coordinates. Since -3 is negative and 4 is positive, the point (-3,4) lies in Quadrant II. .
What is the y-intercept of the line described by y = 2x + 5?
(0,5)
(2,5)
(5,0)
( - 5,0)
In slope-intercept form y = mx + b, b is the y-intercept. Here b = 5, so the y-intercept is the point (0,5). This is where the line crosses the y-axis. .
What is the x-intercept of the line 3x - 6y = 12?
(4,0)
(0,2)
(0, - 2)
( - 4,0)
To find the x-intercept, set y=0: 3x - 6(0) = 12, so x = 4. The intercept is (4,0). This is where the line meets the x-axis. .
The point (-5, -2) is in which quadrant?
Quadrant I
Quadrant IV
Quadrant III
Quadrant II
Quadrant III contains points with both coordinates negative. Here x = - 5 and y = - 2 are both negative, so the point is in Quadrant III. .
What is the slope of a horizontal line?
-1
Undefined
0
1
A horizontal line has no rise as you move along the x-axis, so the change in y is zero. Slope m = ?y/?x = 0. .
What is the slope of a vertical line?
0
Undefined
1
- 1
For a vertical line, ?x = 0, which makes ?y/?x undefined. Thus the slope of any vertical line is undefined. .
Which point is plotted at (4, 0)?
4 units up on the y-axis
4 units right of the origin on the x-axis
4 units down on the y-axis
4 units left on the x-axis
The x-coordinate 4 means 4 units to the right of the origin, and y = 0 means it lies on the x-axis. Thus the point is 4 units right on the x-axis. .
What is the distance from the origin to the point (3, 4)?
5
1
25
7
Use the distance formula: d = ?(x² + y²) = ?(3² + 4²) = ?(9 + 16) = ?25 = 5. This gives the straight-line distance from the origin. .
What is the midpoint between (0, 0) and (4, 2)?
(1, 1)
(4, 1)
(2, 2)
(2, 1)
The midpoint formula is ((x? + x?)/2, (y? + y?)/2). Here it is ((0+4)/2, (0+2)/2) = (2,1). .
In the point (-2, 3), what is the x-coordinate?
-2
2
3
-3
By convention, the first number in an ordered pair (x, y) is the x-coordinate. Here that value is -2. .
Which line has a slope of 0?
x = 3
y = 3
y = -x
x + y = 3
The equation y = 3 is horizontal, meaning ?y = 0, giving a slope of 0. Vertical lines have undefined slope. .
What is the domain of the discrete set of points {(-1, 2), (0, 5), (3, -1)}?
{2, 0, -1}
{-1, 5, 3}
{-1, 0, 3}
{2, 5, -1}
Domain is the set of all x-values in the relation: -1, 0, and 3. The y-values are part of the range, not the domain. .
What is the range of the discrete set {(-1, 2), (0, 5), (3, -1)}?
{-1, 0, 3}
{-1, 5, 3}
{2, -1, 3}
{2, 5, -1}
Range is the set of all y-values: 2, 5, and -1. These are distinct outputs of the relation. .
What does an open circle on a graph indicate?
The point is not included in the graph
A discontinuity to the left
A discontinuity to the right
The point is included
An open circle shows that the function value at that x is not included, indicating a hole or excluded endpoint. Closed circles indicate inclusion. .
What is the y-coordinate of the point (6, -7)?
-7
7
-6
6
In the ordered pair (x, y), the second value is the y-coordinate. Here that value is -7. .
What is the slope of the line passing through (2, 3) and (5, 11)?
8/3
- 8/3
4/3
3/8
Slope m = (y? - y?)/(x? - x?) = (11 - 3)/(5 - 2) = 8/3. This measures the rise over run between the two points. .
Which equation represents a line with slope 2 passing through (0, 1)?
y = - 2x + 1
y = x + 2
y = 2x + 1
y = 2x - 1
The slope-intercept form is y = mx + b. With m=2 and passing through (0,1), b=1, giving y=2x+1. .
What is the equation of the vertical line through x = 4?
y = 4x
y = 4
x = 4
4x + y = 0
A vertical line through x = k is written as x = k. It includes all points whose x-coordinate is 4. .
Which relation is not a function? { (1,2), (1,3), (2,4) }
Because x=1 maps to two different y-values
It is a function
Because y-values repeat
Because x=2 only appears once
A function assigns each x-value exactly one y-value. Here x=1 maps to both 2 and 3, violating the definition. Function definition.
If f(x) = x², what is f( - 3)?
- 9
9
- 6
6
Plugging in x = - 3 gives f( - 3) = ( - 3)² = 9. Squaring a negative yields a positive result. Function evaluation.
What is the graph shape of y = |x|?
A V-shape
A U-shape
A straight line
A downward parabola
The absolute value function reflects negative x-values up, forming a V shape with its vertex at the origin. .
What is the domain of f(x) = ?(x + 2)?
x ? 2
x > 0
All real x
x ? - 2
The radicand x + 2 must be ? 0, so x ? - 2. Negative radicands yield non-real outputs. .
What is the range of y = x²?
y ? 0
-? < y < ?
All real y
y ? 0
Since squaring any real x yields a non-negative result, y must be ? 0. There are no negative outputs. .
Convert 2x + 3y = 6 into slope-intercept form.
y = (2/3)x + 6
y = (3/2)x + 6
y = - 2x + 6
y = - (2/3)x + 2
Solve for y: 3y = - 2x + 6, then y = - (2/3)x + 2. This expresses the line as y=mx+b. .
If y = 2x + 4, what is x when y = 0?
0
4
2
- 2
Set 0 = 2x + 4, so 2x = - 4, giving x = - 2. This is the x-intercept of the line. .
Which of these points lies on the line y = - x + 5?
(5, 1)
(0, 0)
(2, 3)
(1, 1)
Plug in x=2: y = - 2 + 5 = 3, so (2,3) satisfies the equation. The others do not satisfy y= - x+5. .
What is the axis of symmetry of the parabola y = (x - 2)² + 1?
y = 1
x = 2
x = 1
y = 2
For y = (x - h)² + k, the axis of symmetry is x = h. Here h=2, so x=2. .
How is the graph of y = 2(x + 3) - 4 shifted from y = 2x?
3 units right and 4 units up
3 units left and 4 units up
3 units right and 4 units down
3 units left and 4 units down
Replacing x with (x+3) shifts the graph left 3; subtracting 4 shifts it down 4. Vertical stretches by factor 2 remain. Function transformations.
What transformation does y = f(x - 2) represent?
Shift left by 2 units
Shift down by 2 units
Shift right by 2 units
Shift up by 2 units
Replacing x with (x - 2) moves the graph to the right by 2 units. This inside shift is opposite in direction. Function shifts.
What is the slope of a line perpendicular to y = 3x + 1?
3
1/3
- 1/3
- 3
Perpendicular slopes are negative reciprocals. The reciprocal of 3 is 1/3, so its negative reciprocal is - 1/3. .
What is the equation of the line parallel to y = 4x - 2 that passes through (1, 3)?
y = - 4x + 3
y = 4x + 1
y = (1/4)x + 3
y = 4x - 1
Parallel lines share slope m=4. Using point-slope form: y - 3 = 4(x - 1) gives y = 4x - 1. .
Find the equation of the perpendicular bisector of the segment joining (0, 0) and (4, 2).
y = (1/2)x + 1
y = - 2x + 5
y = 2x + 1
y = - (1/2)x + 2
Midpoint is (2,1). Original slope is 2/4=1/2, so perpendicular slope is - 2. Using y - 1 = - 2(x - 2) gives y = - 2x +5. .
What is the distance between (1, 2) and (4, 6)?
5
?13
7
?29
Distance formula: ?[(4 - 1)² + (6 - 2)²] = ?(9 + 16) = ?25 = 5. .
Which three points are collinear? (1,2), (2,4), (3,6), (4,7)
(2,4), (3,6), (4,7)
(1,2), (2,4), (4,7)
(1,2), (2,4), (3,6)
(1,2), (3,6), (4,7)
Slope between (1,2) and (2,4) is 2, and between (2,4) and (3,6) is also 2, so they lie on the same line. The others do not share a constant slope. .
Find the equation of the line through (2, 3) and (5, 7) in slope-intercept form.
y = (3/4)x + 1
y = 3x - 3
y = (4/3)x - 1/3
y = (4/3)x + 1/3
Slope m = (7 - 3)/(5 - 2) = 4/3. Use y - 3 = (4/3)(x - 2), giving y = (4/3)x + 1/3. .
Identify the vertex of the parabola y = 2(x + 1)² - 3.
(1, 3)
(-1, -3)
(0, -1)
(-2, -3)
In y = a(x - h)² + k form, (h, k) is the vertex. Here x+1 means h = - 1, k = - 3. .
Is f(x) = - x² + 4x + 1 concave up or down?
Upward
Neither
Both
Downward
The leading coefficient is - 1, which is negative, so the parabola opens downward. .
What are the x-intercepts of y = x² - 5x + 6?
x = 1 and x = 6
x = 2 and x = 3
x = 5 and x = 6
x = - 2 and x = - 3
Factor to (x - 2)(x - 3)=0, so x = 2 or 3. These are the points where the graph crosses the x-axis. .
For the rational function y = (x + 2)/(x - 3), what is the vertical asymptote?
x = - 2
y = 3
x = 3
y = - 3
Vertical asymptotes occur where the denominator is zero, so x - 3=0 gives x=3. The graph approaches this line without touching. .
What is the horizontal asymptote of y = (x + 2)/(x - 3)?
x = 1
y = 3
y = - 1
y = 1
For rational functions with equal degree numerator and denominator, the horizontal asymptote is the ratio of leading coefficients: 1/1 = 1. .
What is the period of y = sin(2x)?
2?
?/2
4?
?
The period of sin(bx) is 2?/b. Here b=2, so the period is 2?/2 = ?. .
What is the domain of y = ln(x)?
x > 0
x ? 0
x < 0
All real x
The natural log function requires x > 0 because ln of zero or negative numbers is undefined in the real number system. .
What is the range of y = e??
y ? 0
y > 0
All real y
y < 0
Exponential functions of the form e? are always positive for real x, so the output y is greater than zero. .
What is the end behavior of y = 1/x as x ? ??
y ? 1
y ? ?
y ? 0
y ? - ?
As x grows very large, 1/x becomes very small, approaching 0. This describes the horizontal asymptote at y=0. .
Which of these is the midpoint formula?
?((x? - x?)²+(y? - y?)²)
((x? + x?)/2, (y? + y?)/2)
(x?y? + x?y?)
(y? - y?)/(x? - x?)
The midpoint of a segment between (x?,y?) and (x?,y?) is the average of the coordinates: ((x?+x?)/2, (y?+y?)/2). .
Given the piecewise function f(x) = { x+2 for x<1; 2x - 1 for x?1 }, what is f(1)?
1
0
3
2
For x ? 1, f(x) = 2x - 1. Substituting x = 1 gives f(1) = 2(1) - 1 = 1. The first piece applies only for x<1. Piecewise functions.
What is the domain of f(x) = 1/(x² - 4)?
x > 2 or x < - 2
All real x
x ? 0
x ? 2 and x ? - 2
The denominator x² - 4 factors to (x - 2)(x+2). Setting it ? 0 excludes x = 2 and x = - 2. .
For f(x) = ?(5 - 2x), what is the domain?
x ? 5
All real x
x ? 2.5
x ? 2.5
Inside the radical 5 - 2x must be ? 0. Solve 5 - 2x ? 0 gives x ? 2.5. Values above 2.5 make the expression negative. .
Simplify f(x) = (x² - 1)/(x - 1) and state any restriction on x.
f(x) = x + 1, no restriction
f(x) = x - 1, x ? 1
f(x) = x - 1, no restriction
f(x) = x + 1, x ? 1
Factor numerator: (x - 1)(x+1)/(x - 1) simplifies to x+1, but x=1 is excluded as it makes the original denominator 0. .
What is the limit of f(x) = (x² - 1)/(x - 1) as x ? 1?
2
1
Undefined
0
After simplification to x+1 (for x?1), the limit as x?1 is 1+1=2. Limits can exist where the function is undefined. .
Given f(x)=2x+3 and g(x)=x², what is (f ? g)(x)?
2x + 3²
(2x+3)²
x² + 3
2x² + 3
Composition f(g(x)) means f(x²) = 2(x²) + 3 = 2x² + 3. Function composition.
Is the piecewise function f(x)= { x+2 for x<1; 2x - 1 for x?1 } continuous at x=1?
No, it has a jump discontinuity
No, infinite discontinuity
No, removable discontinuity
Yes, continuous
Left-hand limit: 1+2=3, right-hand: 2(1) - 1=1, and f(1)=1. Limits differ, so there's a jump discontinuity at x=1. .
0
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Study Outcomes

  1. Interpret Coordinate Planes -

    After completing this graphs practice, learners will be able to identify axes, quadrants, and exact grid points on a coordinate plane, ensuring clarity when reading and plotting data.

  2. Plot Points and Relations -

    Participants will practice plotting ordered pairs and mapping linear and non-linear relations accurately, strengthening their skills for any graphing quiz or real-world application.

  3. Analyze Linear Equations -

    Users will learn to determine slope and intercepts from equations, sketch corresponding lines on the coordinate grid, and interpret their geometric meaning.

  4. Identify Curve Intersections -

    Through targeted questions in the graphing test, learners will recognize and compute intersection points of functions, improving problem-solving strategies.

  5. Evaluate Function Behavior -

    Readers will assess increasing, decreasing, and constant intervals of curves, gaining insight into function trends and critical values.

  6. Self-Assess Graphing Skills -

    By reviewing answers and explanations, participants will pinpoint strengths and areas for improvement, boosting confidence for future graphing quizzes and tests.

Cheat Sheet

  1. Coordinate Plane Fundamentals -

    Master the layout of the x- and y-axes, origin, and four quadrants to accurately plot ordered pairs like (3, - 2). According to Khan Academy, visualizing how positive and negative regions mirror each other helps reduce sign errors during graphs practice. Remember the mnemonic "All Students Take Calculus" to recall sign patterns in Quadrants I - IV.

  2. Slope and Slope-Intercept Form -

    Understand that slope (m) = (y₂ - y₝)/(x₂ - x₝) measures a line's steepness, and the equation y = mx + b (slope-intercept) shows rise over run plus y-intercept. MIT OpenCourseWare emphasizes plugging in two known points to compute m, then solving for b to graph any line quickly. For example, points (1,2) and (3,6) give m = 2 and b = 0, so y = 2x.

  3. Point-Slope Form for Quick Graphing -

    The point-slope formula y - y₝ = m(x - x₝) lets you sketch a line when you know one point and the slope - ideal for timed graphing quiz sections. As per Purdue University's math tutorials, this form prevents recalculating intercepts: insert (x₝,y₝) and m directly. For instance, with m = - 1 through (2,3), you get y - 3 = - 1(x - 2).

  4. Finding Intercepts and Zeroes -

    Set y = 0 to find the x-intercept and x = 0 for the y-intercept; plotting these two points yields a precise line in any graphing test. The University of California's online notes recommend checking both intercepts for linear equations like 2x + 3y = 6, which gives x-intercept 3 and y-intercept 2. This method also helps verify where functions cross the axes.

  5. Distance and Midpoint Formulas -

    Use the distance formula √[(x₂ - x₝)² + (y₂ - y₝)²] to calculate the length between two points, and the midpoint formula ((x₝+x₂)/2, (y₝+y₂)/2) to find a segment's center. These tools, highlighted by the National Council of Teachers of Mathematics, are crucial when interpreting graphs or checking if a plotted point bisects a line. For example, the midpoint between (1,4) and (5,8) is (3,6).

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