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Find a formula for the quadratic function whose graph has vertex (−2,−1) and y-intercept 6.

 Jul 1, 2021
 #1
avatar+35059 
+2

Vertex form of the parabola

y = a ( x+2)^2  -1          sub in  0,6    to calculate 'a'

6 = 4a - 1

7 = 4a

a = 7/4

 

 

y = 7/4 (x+2)^2 -1        expand and simplify to the form you need (quadratic)

 

y = 7/4 (x^2 + 4x + 4) -1

 

y = 7/4 x^2 + 7x +6

 Jul 1, 2021
edited by Guest  Jul 1, 2021
 #2
avatar+121048 
+1

A  y intercept   of  6 means   that   (0,6)  is on the  graph

 

We  have  the  form

 

y = a ( x - h)^2  +  k           where    x = 0   y = 6     h = -2    and  k  = -1

 

We need to solve for  a

 

6 = a ( 0 + 2)^2   -  1

6 + 1  =  4a

7  = 4a

a  =  7/4

 

The  function  is

 

y = (7/4) ( x + 2)^2  -  1

y = (7/4)  ( x^2  + 4x  + 4 )  - 1

y = (7/4) x^2   + 7x  +  7   -1

y = (7/4)x^2  + 7x  + 6

 

 

cool cool cool

 Jul 1, 2021

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