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Finding Equations of Parabolas Given the Vertex.

 

it becomes challenging when the vertex of a parabola is given though, the rest of the equation is still to be discovered. 

 

However, more information is needed to find the equation of a parabola. The equation cannot simply be solved with the vertex of the parabola. 

 

For example; the vertex of a parabola is at (4,8) one of the x- intercepts is (3,0). What is the equation of the parabola?

 

The formula or equation used to find the equation of parabola is y=a (x-h)2+ k

 

• First, put in your vertex in the equation. Which means putting in you x- coordinate as the h value because it tells us the horizontal shift in the parabola and the y- coordinate in the other because it tells us the vertical shift of the parabola.

y=a (x-4)2 +8 

 

• Second, Now you can substitute your x/y intercept in the equation to find the "a" value. 

y= a (x-4)2 + 8

0= a (3-4)2+8

0=a (-1)2 + 8

0-8 = 1a

-8=1a

-8  = 1a

1   =   1

-8= a

 

• Now, you have found the a value, you have the equation of parabola. 

 

y = -8 (x-4) + 8  is the equation of the parabola for this example. 

 

 

X- intercepts or Zeroes. 

 

 In order to find your x intercepts you need the vertex of the parabola. The vertex will be at the point (h,k). 

• Find the y- intercept, let x = 0 and solve for y. 

• To solve for x intercepts, zeroes, or roots ,let y= 0 and solve for x. 

 

Example; y= 3 (x+1)^2 - 5

 

Vertex (h,k): (-1, -5)

 

Find for y- intercept, let x= 0 and solve for y. 

 

y= 3 (0+1)^2 -5

y=3 (1)^2 -5

y= 3 (1)-5

y= 3-5

y= -2 

 

y- intercept is at (0, -2) 

 

Find the x intercepts, let y = 0 and solve for x. 

 

0= 3 (x+1)^2 -5

5= 3 (x+1)^2

3          3                          * to get rid of the square you do the opposite meaning square root. 

)15  = x+1                       * move 1 to the other side. 

  3 

)15 +-1 = x 

   3

 

x = 0.29 (0.29,0)

 

x = - 2.29 (-2.29,0)