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
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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.
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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.
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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.
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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.
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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.
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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
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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).
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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.
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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.
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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.
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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.
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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.
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