Intersection of Graphs and the Square Root Graph: PECTAA Class 11 Maths, Punjab Board

Two graphs on one sheet of graph paper tell you more than either does alone. Wherever they cross, one pair of x and y values works for both rules at once. Finding that crossing point by drawing is what section 2.2 of the book is about.

Unit 2 of the PECTAA Mathematics 11 book is Functions and Graphs. After the concept of a function (section 2.1), section 2.2 finds intersecting points graphically, and section 2.3 draws the graph of the square root function. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Point of intersection (section 2.2). It is a point where two or more graphs meet on the coordinate plane. Because the point lies on both graphs, its coordinates are a solution of both equations together.

A line and the coordinate axes (section 2.2.1). A linear function has highest power of x equal to 1, so its graph is a straight line. It crosses the y-axis where x = 0 and the x-axis where y = 0. Make a short table of values, plot the points, join them, and read off the two crossings.

Two lines (section 2.2.2). Two linear functions meet where their graphs cross. At that point both functions have the same x value and the same y value. Draw both lines from tables of values on the same axes and read the crossing.

A line and a parabola (section 2.2.3). A line can meet a parabola in:

• two points: the system has two solutions;
• one point: a single solution, and the line may be a tangent to the parabola;
• no point: no solution, so the graphs do not meet.

Square root function (section 2.3). For y = 2√x + 1, the square root of a negative number is not real, so the domain is x ≥ 0. The smallest value of √x is 0, so the smallest y is 1 and the range is y ≥ 1. The graph starts at (0, 1) and rises more and more slowly.

Key points

• An intersection point satisfies both equations. Always check it in both.
• To draw a parabola y = ax² + bx + c well, find its y-intercept, its x-intercepts and its vertex (h, k), where h = −b/(2a) and k is the value of y at x = h.
• A graphical answer is read from the drawing, so it can be approximate. Substitution confirms it.
• For a square root graph, settle the domain first, then pick x values inside it.

Worked examples

Example 1 (book Example 7). Find graphically where y = 2x + 6 meets the coordinate axes.

1. Put x = 0: y = 6. The line meets the y-axis at (0, 6).
2. Put y = 0: 2x + 6 = 0, so x = −3. The line meets the x-axis at (−3, 0).
3. A few more table values, such as (−1, 4) and (1, 8), confirm the straight line through these two points.

Example 2 (book Example 8). Find graphically the point of intersection of y = 3x + 2 and y = −x + 6.

1. Table for y = 3x + 2: x = 0 gives 2, x = 1 gives 5, x = 2 gives 8.
2. Table for y = −x + 6: x = 0 gives 6, x = 1 gives 5, x = 2 gives 4.
3. Both tables give y = 5 at x = 1. Plotting confirms the lines cross at (1, 5).
4. Check: 3(1) + 2 = 5 and −1 + 6 = 5.

Example 3 (book Example 9). Solve y = −x + 3 and y = x² − 6x + 3 graphically.

1. The line y = −x + 3 has x-intercept (3, 0) and y-intercept (0, 3).
2. For the parabola, x = 0 gives the y-intercept (0, 3).
3. For its x-intercepts, solve 0 = x² − 6x + 3: x = (6 ± √(36 − 12))/2 = 3 ± √6, about 0.6 and 5.4. So (0.6, 0) and (5.4, 0).
4. Vertex: h = −(−6)/(2 × 1) = 3 and k = 9 − 18 + 3 = −6, so the vertex is (3, −6).
5. Plotting both graphs, they meet at (0, 3) and (5, −2).
6. Check (5, −2): −5 + 3 = −2 and 25 − 30 + 3 = −2. Two points, so two solutions.

Example 4 (book Example 10). Graph y = 2√x + 1.

1. Domain x ≥ 0, range y ≥ 1.
2. Table: x = 0 gives 1, x = 1 gives 3, x = 2 gives about 3.8, x = 4 gives 5, x = 9 gives 7.
3. Plot and join with a smooth curve starting at (0, 1).

Common mistakes in the exam

1. Reading the crossing without checking. Substitute the point into both equations.
2. Mixing up intercepts. The x-intercept has y = 0; the y-intercept has x = 0.
3. Wrong vertex sign. For x² − 6x + 3, h = −(−6)/2 = +3, not −3.
4. Plotting a parabola from two or three points. Use the intercepts, the vertex and points on both sides of it.
5. Using negative x in y = 2√x + 1. Those values are outside the domain.
6. Calling one shared point "no solution". One shared point is one solution; the line is a tangent.

Exam technique

• Draw neat tables of values before plotting, and label each graph with its equation.
• Choose a scale that keeps every intercept and the vertex on the page.
• Write the answer as coordinates, for example (1, 5), and add a one-line substitution check.
• For square root graphs, write the domain and range before the table.
• Check your board's notice for the sections in your paper.

Practice questions

1. Where does y = 2x − 4 meet the x-axis? Answer: y = 0 gives x = 2, so at (2, 0).
2. Where does y = 2x − 4 meet the y-axis? Answer: x = 0 gives y = −4, so at (0, −4).
3. Find the point where y = x + 1 and y = −x + 5 meet. Answer: x + 1 = −x + 5 gives x = 2, y = 3. The point is (2, 3).
4. Find the vertex of y = x² − 4x + 1. Answer: h = 2, k = 4 − 8 + 1 = −3. The vertex is (2, −3).
5. State the domain and range of y = √x + 2. Answer: Domain x ≥ 0, range y ≥ 2.

Ready to practise this? Continue in 1st Year Mathematics.