how to get a parabola equation from vertex and x intercepts

Question

I need to find the equation for a parabola  with a

- vertex: 3, -4

- x intercepts: 1,0 and 5,0

 

Many Thanks

 

Answer 1

if y=ax²+bx+c is the parabola, when you set it equal to 0 gives you the x intercepts. So if you can write the quadratic equation that has answers 5 and 1, you will have a good start to write the parabola.

So if x=5 and x=1 are the roots, that comes from

x-5=0 and x-1=0    which comes from

(x-5)(x-1)=0    multiplying out becomes

x²-6x+5=0

 

I can multiply both sides by any number and 1 and 5 will still be the roots. In terms of a parabola, there are an infinite number of parabolas that have these 2 intercepts.

So I'm going to write the parabola as y=a(x²-6x+5)...now this general form of a parabola with x intercepts 1 and 5 must also pass through (3,-4)....so I can plug this in and find out "a".

 

So -4=a(3²-6*3+5).........-4=a(9-18+5)......-4 = a*-4.......so a=1....so your parabola is y=x²-6x+5.

 

Hope this helped.

Answer 2

Use the "vertex form" of the quadratic equation:

 

     y = a(x-h)² + k

 

where (h,k) is the location of the vertex.  In this problem (h,k) is given as (3,-4) so:

 

     y = a(x-3)² - 4

 

To find a, plug in the values from either known point (x-intercepts).  I'll use (1,0):

 

     0 = a(1-3)² - 4

     0 = a(-2)² - 4

     0 = 4a - 4

     4 = 4a

     1 = a

 

So the final equation is:

 

    y = (x-3)² - 4

 

You can convert it to standard form (y = ax²+bx+c) by multiplying out the (x-3)² term and combining like terms:

 

     y = x² - 6x + 9 - 4

     y = x² - 6x + 5

 

--------

 

You can also solve it using the factored form of the quadratic equation, y = a(x-m)(x-n).  Since the x-intercepts are 1 and 5, the factored form is:

 

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

 

To find the value of a, plug in the (x,y) coordinates of the vertex, (3,-4):

 

-4 = a(3-1)(3-5)

-4 = a(2)(-2)

-4 = -4a

1 = a

 

So the factored form is:

 

y = (x-1)(x-5)