FINDING THE EQUATION OF A PARABOLA

The graphical form of a quadratic function will be a parabola. The quadratic equation can be converted into different forms.

  • Standard form
  • Vertex form
  • Factored form

Finding Equation in Standard Form

The quadratic equation which is in the form of

y = ax2 + bx + c

will form a parabola which opens up or opens down.

  • If a < 0, then the parabola opens down.
  • If a > 0, then the parabola opens up.

Finding Equation in Vertex Form

The quadratic equation which is in the form of

y = a(x - h)2 + k

Here (h, k) is the vertex.

Equations in Factored Form

The factored form of a parabola with zeroes r and s.

y = a (x - r) (x - s)

Equation of the Parabola Passing Through the Given Points

Problem 1 :

Find a quadratic model for each set of values.

(–1, 1), (1, 1), (3, 9)

Solution :

Let the quadratic model be y = ax2 + bx + c

The parabola passes through the given points, so we can apply the points one by one.

It passes through (-1, 1).

1 = a(-1)2 + b(-1) + c

1 = a - b + c

a - b + c = 1 -----------(1)

It passes through (1, 1).

1 = a(1)2 + b(1) + c

1 = a + b + c

a + b + c = 1 -----------(2)

It passes through (3, 9).

9 = a(3)2 + b(3) + c

9 = 9a + 3b + c

9a + 3b + c = 9 -----------(3)

(1) + (2)

a - b + c + a + b + c = 1 + 1

2a + 2c = 2

a + c = 1 ----(4)

Multiply the first equation and subtract (3), we get

3a - 3b + 3c + 9a + 3b + c = 3+9

12a + 4c = 12 ----(5)

4(4) - (5)

4a + 4c - 12a - 4c = 4 - 12

-8a = -8

a = 1

Applying a = 1 in (4), we get

1 + c = 1

c = 0

When c = 0 and a = 1

1 + b + 0 = 1

b = 0

So, the required equation of parabola is y = x2

Finding Equation in Vertex form

Problem 2 :

Find the quadratic function with the given vertex and point. 

Vertex (2, 0) passing through (1, 3).

Solution :

y = a(x - h)2 + k

(h, k) ==>  (2, 0)

y = a(x - 2)2 + 0

y = a(x - 2)2 ----(1)

The parabola is passing through the point (1, 3).

3 = a(1 - 2)2

3 = a(-1)2

3 = a

Applying the value of a in (1), we get

y = 3(x - 2)2

Equation of parabola from zeroes or x-intercepts

Problem 3 :

Determine the equation of the parabola whose zeroes are 4 and -5 which passes through (2, -42).

Solution :

Zeroes are 4 and -5.

x = 4 and x = -5

Product of factors :

(x - 4)(x + 5)

Writing it as factored form, we get

y = a (x - 4)(x + 5) ------(1)

The parabola passes through the point (2, -42).

-42 = a (2 - 4)(2 + 5)

-42 = a (-2)(7)

a = 42/14

a = 3

By applying the value of a in (1), we get

y = 3(x - 4)(x + 5)

We can convert this into standard form or vertex form. We can check the answer using graphing calculator.

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