Problem 1:
If
f(2)=f(4)=0, f(3)=−2
x = 2 and x = 4 are roots and the vertex lies at (3, -2)......so we know that
0 = a(2 - 3)^2 - 2 simplify
0 = a (-1)^2 - 2
0 = a - 2 → a = 2
So...our function is f(x) = 2(x - 3)^2 - 2
Check that x = 4 produces a zero
2(4 - 3)^2 - 2 = 2(1)^2 - 2 = 2 - 2 = 0
Problem 2:
If
Zeroes at 1± sqrt(2), y-intercept at –4
The x value of the vertex will be the average of the sum of the roots =
[ 1 + sqrt (2) + 1 - sqrt(2) ] / 2 = 2 / 2 = 1
And the x coordinate of the vertex = -B/ [ 2A] implies that
-B/ [2A] = 1 which implies that
-B = 2A which implies that
B = -2A
So we have the following :
y = Ax^2 + Bx + C and substituting for B
y = Ax^2 - (2A)x + C
And since the y intercept is at - 4 , then
-4 = A(0)^ -2A(0) + C
-4 = 0 + 0 + C
-4 = C
So we have that
y = Ax^2 -2Ax - 4
And we know that
0 = A(1 + sqrt(2))^2 - 2A(1 + sqrt(2)) - 4
4 = A ( 1 + 2sqrt(2) + 2) - 2A - 2Asqrt(2)
4 = A + 2Asqrt(2) + 2A - 2A - 2Asqrt(2)
4 = A .... and B = -2A = -8
So we have
y = 4x^2 - 8x - 4
Here's a graph that confirms this : https://www.desmos.com/calculator/ba8scvgskt