What is the vertex form of a parabola given vertex (41,71) & zeros (0,0) (82,0)?

1 Answer
Mar 1, 2016

The vertex form would be # -71/1681(x-41)^2+71#

Explanation:

The equation for vertex form is given by:
# f(x) = a(x-h)^2 + k# , where the vertex is located at point #(h,k) #

So,substituting the vertex #(41,71)# at #(0,0)#,we get,

#f(x) = a(x-h)^2 + k#

#0=a(0-41)^2+71#

#0 = a(-41)^2 + 71 #

#0 = 1681a + 71#

#a = -71/1681#

So the vertex form would be
#f(x) = -71/1681(x-41)^2+71#.