A parabola with equation $y=ax^2+bx+c$ has a vertical line of symmetry at $x=2$ and goes through the two points $(1,1)$ and $(4,-1)$. The quadratic $ax^2 + bx

Since 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 )

Let $b$ and $c$ be constants such that the quadratic \(-2x^2 +bx +c\) has roots \(3+\sqrt{5}\) and \(3-\sqrt{5}\). Find the vertex of the graph of the equation \(y=-2x^2 + bx + c\).

The sum of the roots of a quadratic  =  -b/a

So ........

[3 + √5[  + [3 - √5]  = -b/a

6  =  -b/-2    →    b = 12

And the product of the roots of a quadratic =  c / a.....so we have that

[3 + √5] [ 3 - √5]  =  c / -2

[9 - 5]   = c /-2

4  = c / -2  

c = -8

So......our quadratic is....

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

And the x  coordinate of the vertex =   -b/2a   =  -12/ [2(-2)] = 12/4 = 3

And the y coordinate is  .......

-2(3)^2 + 12(3) -  8

-2(9) + 36 - 8

-18 + 36 - 8

10

So....rhe vertex is  (3, 10)

Here's the graph : https://www.desmos.com/calculator/aqwwi79n10