Math Problem Statement

Find the equation of the parabola passing through (1,0),(2,12) and (-2,-12)

Solution

To find the equation of a parabola passing through three points, we can use the general form of the equation of a parabola:

y=ax2+bx+cy = ax^2 + bx + c

Given points: (1,0)(1, 0), (2,12)(2, 12), and (2,12)(-2, -12).

We'll substitute each point into the equation to form a system of linear equations.

Step 1: Substitute the points into the parabola equation

  1. For (1,0)(1, 0): 0=a(1)2+b(1)+c    a+b+c=00 = a(1)^2 + b(1) + c \implies a + b + c = 0

  2. For (2,12)(2, 12): 12=a(2)2+b(2)+c    4a+2b+c=1212 = a(2)^2 + b(2) + c \implies 4a + 2b + c = 12

  3. For (2,12)(-2, -12): 12=a(2)2+b(2)+c    4a2b+c=12-12 = a(-2)^2 + b(-2) + c \implies 4a - 2b + c = -12

Step 2: Solve the system of equations

We have the following system of equations:

  1. a+b+c=0a + b + c = 0
  2. 4a+2b+c=124a + 2b + c = 12
  3. 4a2b+c=124a - 2b + c = -12

Let's subtract equation 1 from equations 2 and 3 to eliminate cc:

  • Subtract equation 1 from equation 2: (4a+2b+c)(a+b+c)=120(4a + 2b + c) - (a + b + c) = 12 - 0 3a+b=12(Equation 4)3a + b = 12 \quad \text{(Equation 4)}

  • Subtract equation 1 from equation 3: (4a2b+c)(a+b+c)=120(4a - 2b + c) - (a + b + c) = -12 - 0 3a3b=12(Equation 5)3a - 3b = -12 \quad \text{(Equation 5)}

Step 3: Solve for aa and bb

From Equation 5: ab=4(Divide by 3)a - b = -4 \quad \text{(Divide by 3)} b=a+4(Equation 6)b = a + 4 \quad \text{(Equation 6)}

Substitute Equation 6 into Equation 4: 3a+(a+4)=123a + (a + 4) = 12 4a+4=124a + 4 = 12 4a=84a = 8 a=2a = 2

Substitute a=2a = 2 into Equation 6: b=2+4=6b = 2 + 4 = 6

Step 4: Solve for cc

Using a=2a = 2 and b=6b = 6 in the first original equation: 2+6+c=02 + 6 + c = 0 c=8c = -8

Final Equation of the Parabola

The equation of the parabola is:

y=2x2+6x8\boxed{y = 2x^2 + 6x - 8}

Would you like more details on this process or have any questions?

5 Related Questions

  1. How can you verify that the parabola passes through the given points?
  2. What are the key features (vertex, axis of symmetry, etc.) of the parabola y=2x2+6x8y = 2x^2 + 6x - 8?
  3. How do you find the focus and directrix of a parabola given its equation?
  4. What is the significance of the discriminant in determining the nature of the roots of a quadratic equation?
  5. How can you derive the vertex form of a parabola from the standard form equation?

Tip

Always double-check your algebraic manipulations when solving systems of equations to avoid errors in the final result.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
System of Equations
Parabolas

Formulas

General form of a parabola: y = ax^2 + bx + c

Theorems

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Suitable Grade Level

High School