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".

HelpBot Jul 28, 2020

#1**+2 **

nce the line of symmetry is x = 2, the x coordinate of the vertex must = 2

And the x coordinate of the vertex is given by : -b/2a.......so we have that

2 = -b/[2a] → -4a = b

And we have these equations

1 = a(1)^2 - 4a(1) + c → 1 = -3a + c (1)

-1 = a(4)^2 - 4a(4) + c → -1 = c (2)

Subbing (2) into (1) we have that 1 = -3a - 1 → 2 = -3a → a = -2/3

Which means that b = -4(-2/3) = 8/3

So....the equation of the quadratic is :

y = (-2/3)x^2 + (8/3)x - 1

And we know that

0 = (-2/3)(√n +2)^2 + (8/3)(√n + 2) - 1 let √n + 2 = m.......and we have that

0 = (-2/3)m^2 + (8/3)m - 1 multiply both sides by 3

0 = -2m^2 + 8m - 3 complete the square on m

3 - 8 = -2 (m^2 - 4m + 4)

-5 = -2 (m - 2)^2

5/2 = (m - 2)^2 take pos/neg square roots

±√[5/2] = m - 2

±√[5/2] + 2 = m back-substituting, we have

±√[5/2] + 2 = √n + 2 and since we want the larger root

√[5/2] = √n

n = 5/2 = 2.5

Here's the graph : https://www.desmos.com/calculator/102umpw19h

The larger root = ( √2.5 + 2 , 0 ) ≈ ( 3.581, 0 )

Guest Jul 28, 2020

#3**+1 **

Close...but the vertex is in the wrong place....

here is a graph:

https://www.desmos.com/calculator/atsyq7hk8i

ElectricPavlov
Jul 28, 2020

#2**+3 **

In vertex form: y = a (x-2)^2 + 4

Sub in the point given:

1 = a(1-2)^2 + 4

1 = a + 4 a = -3

y = -3(x-2)^2 + 4 expand

y = -3x^2 +12x-8

ElectricPavlov Jul 28, 2020