Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c". Thanks!

EvanWei123 Dec 30, 2019

#1**+1 **

Here is a way:

Vertex value of 'x' is at -b/2a soooo -b/2a = 2 (1)

and when x = 2 f(x)=4

4 = a(2^2)+b(2) + c or 4 = 4a+2b+c (2)

and you are given point 1,1 1 = a(1^2)+b(1)+c or 1 = a+b+c (3) 3 equations , 3 unknowns

re-arrange (1) : -b/2 = 2a multiply by 2

-b =4a sub into (2)

4 = -b+2b+c 4 = b+c c = 4-b

Take (3) and multiply by 4 4 = 4a + 4b + 4c

4 = -b+4b + 4(4-b) so b = 12 then -b/2a=2 means a = -3 then a + b + c =1 means c = -8

ElectricPavlov Dec 30, 2019