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Equation of a Parabola

Equation of a Parabola. The equation for a parabola can come in 2 forms: General form f(x) = ax 2 + bx + c Standard form f(x) = a(x – h) 2 + k. Each form has its advantages. General form is simpler and therefore a little easier to work with when substituting values.

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Equation of a Parabola

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  1. Equation of a Parabola

  2. The equation for a parabola can come in 2 forms: General form f(x) = ax2 + bx + c Standard form f(x) = a(x – h)2 + k Each form has its advantages. General form is simpler and therefore a little easier to work with when substituting values. Standard form has the advantage of easily identifying the vertex – the most important point of the parabola.

  3. Vertex: Identifying the Vertex To determine the vertex from general form, you can use a formula to determine the x-value and another one to determine its image. f(x) = ax2 + bx + c ∆ = b2 – 4ac To determine the vertex from standard form, you use the parameters h and k presented in the equation. f(x) = a(x - h)2 + k Vertex: (h,k)

  4. Identifying the Axis of Symmetry After determining the vertex, the axis of symmetry is determined by using the equation: x = ‘x-value of the vertex’ This is true whether the parabola is presented in general form or standard form. Therefore, from general form, the axis of symmetry is: Therefore, from standard form, the axis of symmetry is: x = h

  5. Identifying the Zeros To determine the zeros from general form, you can use the quadratic formula. f(x) = ax2 + bx + c ∆ = b2 – 4ac This formula can yield either 0, 1 or 2 values depending on the value of the discriminant. These values manifest themselves as x-intercepts on the graph. To determine the zeros from standard form, it is best to transform the equation to general form and use the method prescribed above. f(x) = a(x - h)2 + k

  6. Transforming from Standard Form to General Form An equation in standard form can be transformed by first using FOIL to multiply the binomial that is squared. f(x) = -2(x - 4)2 + 8 f(x) = -2(x – 4)(x – 4) + 8 = -2(x2 – 4x – 4x + 16) + 8 = -2x2 + 16x - 32 + 8 f(x) = -2x2 + 16x - 24 a = -2; b = 16; c = -24 ∆ = b2 – 4ac = 162 – 4(-2)(-24) = 256 - 192 = 64

  7. Identifying the Y-intercept To determine the y-intercept from general form, you can use parameter c. f(x) = ax2 + bx + c Vertex: (0,c) To determine the y-intercept from standard form, you find f(0). f(x) = -2(x - 4)2 + 8 f(0) = -2(0 - 4)2 + 8 = -2(16) + 8 = -32 + 8 = -24 Vertex: (0,-24)

  8. Transforming from General Form to Standard Form Just as it is possible to transform an equation in standard form to general form, it is possible to transform it the other way. Before looking at this process, it is necessary to review some concepts.

  9. x2 + 2x + 1 x2 - 4x + 4 x2 + 6x + 9 x2 + 8x + 16 x2 - 10x + 25 x2 + 12x + 36 x2 + 14x + 49 x2 - 16x + 64 x2 + 18x + 81 x2 - 20x + 100 To convert from general form to standard, we must use a technique called ‘Competing the Square’. This refers to creating a perfect square trinomial and accommodating the changes within an equation to make that trinomial. Some examples of perfect square trinomials are: x2+2x + 1 Notice that the sign of the middle term can be positive or negative. There is a relationship between the coefficient of the middle term and the last term: x2- 4x + 4 21 -44 x2+14x + 49 1449 x2-20x + 100 -20100

  10. Divide the coefficient of the middle term by 2. Square the result of this division. Divide the coefficient of the middle term by 2. Square the result of this division. x2 -x + Divide the coefficient of the middle term by 2. Square the result of this. The relationship between these terms of the trinomial is what makes the trinomial aperfect square trinomial. We can use this relationship to create a perfect square trinomial. If we know the value of the middle term, we can determine the third term. x2 + 32x + ___ x2 +32x+ 256 x2 +7x +____

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