Free graphing resources · Original practice
Graph four lines from standard-form equations by finding x- and y-intercepts. Keep zero and negative intercepts as valid coordinates.
Graph four lines from standard-form equations by finding x- and y-intercepts. Keep zero and negative intercepts as valid coordinates.
Choose Letter or A4, then download a student sheet first. Keep the separate reference or answer file closed until you have attempted the case. Every worksheet, SVG and answer is free without signup.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Original worksheet collection. Letter or A4 PDF and editable SVG.
Complete free PDF pack.
| Case | Focus |
|---|---|
| ST01 | Positive intercepts |
| ST02 | A falling line |
| ST03 | A line through the origin |
| ST04 | Mixed signs |
All examples use original fictional data and coordinate units. They are practice material, not measured classroom results, published questions or evidence about real people.
ST01: Positive intercepts
| Check | Worked result |
|---|---|
| 1 | x-intercept 6; y-intercept 4. |
| 2 | Equation 2x + (3)y = 12. |
| 3 | Use two intercepts to draw the line. |
The first case shows the checking pattern. Individual answer files contain the completed graph and result for every case.
Begin by identifying the specific linear equation provided in the resource, such as the example with coefficients two and three equaling twelve. To locate the x-intercept, substitute zero for the variable y in the equation. Solve the resulting arithmetic expression to determine the numerical value of x. This calculated value represents the horizontal coordinate where the line crosses the axis. Record this number precisely as it will serve as one of the two anchor points needed for drawing. Ensure you use the exact equation supplied in the listed file rather than estimating from any visual representation of the line.
Next, determine the vertical crossing point by setting the variable x to zero within the same linear equation. For instance, in the case where coefficients are three and negative two equaling six, substitute zero for x and solve for y. The resulting value is the y-intercept coordinate. This step is critical because it provides the second anchor point required to define the line's position on the grid. Remember that a zero intercept is a valid result, particularly in cases where the line passes through the origin. Use the supplied values to calculate this coordinate accurately without relying on visual estimation.
With both intercept coordinates calculated, locate these two points on the provided listed coordinate grid. Plot the x-intercept on the horizontal axis and the y-intercept on the vertical axis. Use a straightedge to connect these two distinct points, extending the line across the grid. The accuracy of the drawn line depends entirely on the precision of the calculated intercepts derived from the exact equation. Do not adjust the line to match a perceived appearance; instead, trust the mathematical coordinates determined by the coefficients. This method ensures the graphical representation aligns perfectly with the algebraic definition provided in the original fictional practice material.
| Watch for | Check instead |
|---|---|
| Estimating the line's position by looking at the printed graph rather than calculating the intercepts from the equation coefficients. | Always calculate the x and y intercepts by substituting zero for the opposite variable in the exact equation provided, then plot those specific coordinates. |
| Assuming a line must cross both axes at non-zero values and discarding a result where the intercept is zero. | Recognize that a zero intercept is valid, especially when the constant term is zero, indicating the line passes through the origin as seen in specific cases. |
| Ignoring negative coefficients in the equation, leading to incorrect signs for the calculated intercept coordinates. | Carefully apply the negative signs from the supplied coefficients A and B when solving for the intercepts to ensure the plotted points are in the correct quadrants. |
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Set y to zero and solve for x. For a y-intercept, set x to zero and solve for y.
Yes. A line through the origin can have both intercepts at (0,0).
Keep the sign when solving. A negative intercept is a valid coordinate result.
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Take 3x - 4y = 12. To find the x-intercept, set y = 0: 3x = 12, so x = 4. The point is (4, 0). To find the y-intercept, set x = 0: -4y = 12, so y = -3. The point is (0, -3). Plot both points and draw the line through them.
The slope follows from the intercepts: from (0, -3) to (4, 0), the rise is 3 and the run is 4, so the slope is 3/4. Because the coefficient of x is positive and the coefficient of y is negative, the line rises from left to right. A quick sign check like this catches many plotting errors before they spread.
Check each intercept by substitution rather than by eye. For (4, 0): 3(4) - 4(0) = 12. For (0, -3): 3(0) - 4(-3) = 12. If either result differs, the intercept is wrong even if the line looks plausible.
Check the third point against the drawn line. If (8, 3) does not lie on your line, either the line is misplaced or the arithmetic is wrong. Recompute before redrawing. Also confirm that the line extends beyond both intercepts and that the axes are labelled with equal spacing, since unequal spacing distorts slope.
Try equations where one intercept is zero, such as 2x + y = 0. Here the x-intercept is (0, 0) and the y-intercept is also (0, 0), so a second point is required: choose x = 1, giving y = -2, and plot (1, -2). A single intercept is not enough to define a line through the origin.
Try an equation with both coefficients negative, such as -2x - 3y = 6. The x-intercept is (-3, 0) and the y-intercept is (0, -2). The line falls from left to right. Comparing this with the previous case helps you see how coefficient signs control direction without memorising rules.
Graph 3x - 4y = 12 and verify the line with a third point.
Set y = 0: 3x = 12, so x = 4, giving (4, 0). Set x = 0: -4y = 12, so y = -3, giving (0, -3). Choose x = 8: 24 - 4y = 12, so -4y = -12 and y = 3, giving (8, 3). Substituting: 3(4) - 4(0) = 12, 3(0) - 4(-3) = 12, and 3(8) - 4(3) = 12. All three points lie on one line with slope 3/4.
Solving for y gives y = (3/4)x - 3, so the slope is positive 3/4 and the line rises as x increases.
That happens when the line passes through the origin. Choose another x value, compute y, and plot that second point before drawing.
Substitute the intercepts and one extra point into the original equation. All three must satisfy it exactly.
