SOLUTION: find the quadratic function whose vertex is (-1,5) and passing through the point (2,7) and graph the function.
need answer before 12/1/08
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Question 170246: find the quadratic function whose vertex is (-1,5) and passing through the point (2,7) and graph the function.
need answer before 12/1/08
Found 2 solutions by nerdybill, scott8148:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Reviewing vertex formula:
http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php
.
Standard "vertex form":
y= a(x-h)^2+k
where
(h,k) is the vertex
.
The problem gives you:
(h,k) = (-1,5)
(x,y) = (2,7)
.
Plug the above into:
y= a(x-h)^2+k
7= a(2-(-1))^2+5
Solve for 'a':
7= a(2+1)^2+5
7= a(3)^2+5
7= 9a+5
2 = 9a
2/9 = a
.
Therefore:
y= a(x-h)^2+k
y= (2/9)(x-(-1))^2+5
y= (2/9)(x+1)^2+5
y= (2/9)(x^2+2x+1)+5
9y= 2(x^2+2x+1)+45
9y= 2x^2+4x+2+45
9y= 2x^2+4x+47
y = (2/9)x^2 + (4/9)x + (47/9)
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
using the vertex form of the equation __ y=a(x+1)^2+5
substituting the point to find a __ 7=a(2+1)^2+5
subtracting 5 __ 2=9a
dividing by 9 __ 2/9=a
y=(2/9)(x+1)^2+5 __ y=(2/9)x^2+(4/9)x+(47/9)
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