Quadratic graphs: solving equations by graph
The graph of a quadratic is a parabola. The x-values where it meets the x-axis are the solutions of the equation.
What to know about the parabola
- The graph of y = ax² + bx + c is a smooth U-shaped curve called a parabola.
- If a > 0 it opens upward. If a < 0 it opens downward.
- It is symmetric about the vertical line x = −b/(2a). Its turning point on this line is the vertex.
- The solutions of ax² + bx + c = 0 are the x-coordinates where the curve meets the x-axis, because y = 0 there.
- Join the plotted points with a smooth free-hand curve, never with a ruler.
Table of values for y = x² − 2x − 3
| x | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| x² | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
| −2x | 4 | 2 | 0 | −2 | −4 | −6 | −8 |
| −3 | −3 | −3 | −3 | −3 | −3 | −3 | −3 |
| y | 5 | 0 | −3 | −4 | −3 | 0 | 5 |
How to draw the graph of y = x² − 2x − 3 and solve x² − 2x − 3 = 0
Use the table of values above and a full graph sheet.
- 1Table of values: take x from −2 to 4. Work out x², −2x and −3 in separate rows, then add each column to get y.
- 2Scale: x-axis 1 cm = 1 unit; y-axis 1 cm = 1 unit. Write the scale in the top right corner of the sheet.
- 3Axes: draw the x-axis and the y-axis through the origin O near the middle of the sheet. Label X′OX and YOY′ and mark the numbers.
- 4Plot the points (−2, 5), (−1, 0), (0, −3), (1, −4), (2, −3), (3, 0) and (4, 5).
- 5Join the points with a smooth free-hand curve. Label it y = x² − 2x − 3.
- 6★Read: the curve meets the x-axis at (−1, 0) and (3, 0).
- 7★Solution: x = −1 and x = 3, so the solution set is {−1, 3}. Check: (−1)² − 2(−1) − 3 = 0 and 3² − 2(3) − 3 = 0.
★ key step
Worked example: solve a second equation with the same graph
Using the graph of y = x² − 2x − 3, solve x² − 3x − 4 = 0.
- Rewrite: x² − 3x − 4 = 0 is the same as x² − 2x − 3 = x + 1.
- So the roots are where the curve y = x² − 2x − 3 meets the line y = x + 1.
- Table for the line: x = −1, 0, 2, 4 gives y = 0, 1, 3, 5.
- Draw the straight line through (−1, 0), (0, 1), (2, 3) and (4, 5) on the same axes, with a ruler.
- The line meets the curve at (−1, 0) and (4, 5).
- Check: (−1)² − 3(−1) − 4 = 0 and 4² − 3(4) − 4 = 0.
x = −1 and x = 4, so the solution set is {−1, 4}
Reading the nature of the solutions from the graph
| What the curve does | Solutions | Example |
|---|---|---|
| Cuts the x-axis at two points | Real and unequal | y = x² − 2x − 3 cuts it at x = −1 and x = 3 |
| Touches the x-axis at one point | Real and equal | y = x² − 4x + 4 touches it at x = 2 |
| Does not meet the x-axis | No real solutions | y = x² + x + 2 stays above the x-axis |
A cricket ball hit high, or water from a garden hose, follows a path close to a parabola. Engineers use the same curve for bridge arches and dish antennas.
1 marks: “The graph of a quadratic touches the x-axis at exactly one point. What is the nature of its roots?” **Real and equal.**
8 marks: “Draw the graph of y = 2x² and hence solve 2x² + x − 6 = 0.” Table for y = 2x², x from −3 to 3: y = 18, 8, 2, 0, 2, 8, 18. Scale: x-axis 1 cm = 1 unit, y-axis 1 cm = 2 units. Draw the parabola. Since 2x² = −x + 6, draw the line y = −x + 6 through (−2, 8), (0, 6) and (2, 4). They meet at x = −2 and x = 3/2. Solution set {−2, 3/2}.
Table, scale, plot, smooth, read: five words for every graph answer. For the roots, look where y = 0.