Simultaneous Equations & Graphs of Polynomials

Simultaneous Equations: One Linear and One Quadratic

This section shows how to solve a system where one equation is linear and the other is quadratic. We solve by setting the two expressions for y equal and then finding x. Finally, we substitute back to get y.

Worked Examples

• Example 1:
  Equations: y = 2x + 3 and y = x² + x + 1
  1. Set 2x + 3 = x² + x + 1. 2. Rearranging gives x² + x + 1 − 2x − 3 = x² − x − 2 = 0. 3. Factor to (x − 2)(x + 1) = 0. 4. So, x = 2 or x = –1. 5. Substitute back: If x = 2, then y = 2×2 + 3 = 7; if x = –1, then y = 2×(–1) + 3 = 1.
• Example 2:
  Equations: y = 3 – x and y = x²
  1. Set 3 – x = x². 2. Rearranging gives x² + x – 3 = 0. 3. Solve using the quadratic formula: x = [–1 ± √(1 + 12)]/2 = [–1 ± √13]/2.
• Example 3:
  Equations: y = x + 2 and y = x² – 2x + 3
  1. Set x + 2 = x² – 2x + 3. 2. Rearranging gives x² – 3x + 1 = 0. 3. Use the quadratic formula: x = [3 ± √(9 − 4)]/2 = [3 ± √5]/2.
• Example 4:
  Equations: y = 4x – 1 and y = x² + 2
  1. Set 4x – 1 = x² + 2. 2. Rearranging gives x² – 4x + 3 = 0. 3. Factor to (x − 1)(x − 3) = 0. 4. So, x = 1 or x = 3. 5. Substitute back: For x = 1, y = 4(1) – 1 = 3; for x = 3, y = 4(3) – 1 = 11.
• Example 5:
  Equations: y = –x + 5 and y = x² – 4x + 4
  1. Set –x + 5 = x² – 4x + 4. 2. Rearranging gives x² – 3x – 1 = 0. 3. Solve with the quadratic formula: x = [3 ± √(9 + 4)]/2 = [3 ± √13]/2.
• Example 6:
  Equations: y = 2x – 3 and y = x² + 1
  1. Set 2x – 3 = x² + 1. 2. Rearranging gives x² – 2x + 4 = 0. 3. The discriminant is negative (4 – 16 = –12), so there are no real solutions.

Graphs of Polynomials of Degree Not Greater Than 3

This section explains how to sketch graphs of quadratic and cubic functions. We identify key features such as the vertex, intercepts, turning points, and end behavior.

Worked Examples

• Example 1 (Quadratic):
  Graph y = x² – 4
  1. The vertex is found by noting the function is in the form x² – 4, so the vertex is at (0, –4). 2. Set y = 0: x² – 4 = 0 gives x = –2 and 2. 3. The y–intercept is at (0, –4).
• Example 2 (Quadratic):
  Graph y = –x² + 2x + 3
  1. Since the coefficient of x² is negative, the parabola opens downward. 2. Find the vertex using x = –b/(2a): here a = –1, b = 2 so x = 1; substitute to get y = 4. 3. Find x–intercepts by solving –x² + 2x + 3 = 0.
• Example 3 (Cubic):
  Graph y = x³ – x² – x + 1
  1. Try to factor or use test points to find approximate intercepts. 2. Identify where the function crosses the x–axis and note the overall rising and falling behavior.
• Example 4 (Cubic):
  Graph y = 2x³
  1. This is a simple cubic that passes through the origin. 2. As x increases, y increases very fast; as x decreases, y decreases very fast.
• Example 5 (Cubic):
  Graph y = x³ – 3x
  1. Factor as x(x² – 3) to get intercepts: x = 0, x = √3, and x = –√3. 2. Sketch the curve showing a change in direction near these intercepts.
• Example 6 (Cubic):
  Graph y = –x³ + 3x²
  1. Factor as –x²(x – 3); the function touches the x–axis at x = 0 (double root) and crosses at x = 3. 2. Sketch the turning points and note that the graph falls as x increases beyond 3.

JAMB CBT Quiz on Simultaneous Equations & Graphs of Polynomials

Total time: 900 seconds

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This topic shows how to solve a system with one linear and one quadratic equation. Example 1: Set 2x + 3 = x² + x + 1, simplify to x² - x - 2 = 0, factor to get x = 2 or -1, then substitute back to get y. Example 2: Set 3 - x = x², rearrange to x² + x - 3 = 0, and solve using the quadratic formula. Example 3: Set x + 2 = x² - 2x + 3, simplify to x² - 3x + 1 = 0, then solve for x. Example 4: Set 4x - 1 = x² + 2, rearrange to x² - 4x + 3 = 0, factor and solve for x, then compute y. Example 5: Set -x + 5 = x² - 4x + 4, simplify to x² - 3x - 1 = 0, then use the quadratic formula. Example 6: Set 2x - 3 = x² + 1, rearrange to x² - 2x + 4 = 0, note the discriminant is negative, so no real solution. This topic covers sketching graphs for quadratic and cubic polynomials (degree ≤ 3). Example 1: For y = x² - 4, the vertex is (0, -4), x-intercepts are -2 and 2. Example 2: For y = -x² + 2x + 3, the parabola opens downward; the vertex is found using x = -b/(2a). Example 3: For y = x³ - x² - x + 1, identify approximate intercepts and turning points. Example 4: For y = 2x³, note that the graph increases very quickly for large |x|. Example 5: For y = x³ - 3x, factor to find zeros at 0 and ±√3, then sketch the curve. Example 6: For y = -x³ + 3x², factor and note the double root at 0, then sketch the turning point at x = 3.