Assistance with this math problem?

a)
y = x^2-5x+6
y-6 = x^2-5x
y-6 = (x-5/2)^2 -25/4
y = (x-5/2)^2 -25/4 +6
y = (x-5/2)^2 - 1/4

Vertex : (5/2, -1/4)
Axis of symmetry is x=5/2

b)
Let x and y be the numbers
x = y+2
xy = 35
y = 35/x
x= y+2
x = 35/x + 2

Multiply both sides by x
x^2 = 35+2x
x^2-2x-35 = 0
(x-7)(x+5) = 0

x=7
y= 5
The numbers are 7 and 5

a.
y = f(x) = x² - 5x + 6
y = [x² - 5x] + 6
y = [x² - 2(5/2)x + (5/2)²] - (5/2)² + 6
y = [x - (5/2)]² - (1/4)

Vertex : (5/2, -1/4)
Axis of symmetry : x - (5/2) = 0


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b.
Let n and n + 2 be the two numbers.

n(n + 2) = 35
n² + 2n - 35 = 0
(n - 5)(n + 7) = 0
n = 5 or n -7
n + 2 = 7 or -5

Hence, the two numbers are 5, 7 or -7, -5

x^2 - 5x + 6
ax^2 + bx + c

axis of symmetry
-b / 2a

-(-5) / (2 * 1)
5/2

x = 5/2
when x = 5/2

y = (5/2)^2 - 5(5/2) + 6
y = 25/4 - 25/2 + 6
y = 25/4 - 50/4 + 24/4
y = -1/4

vertex (5/2, -1/4)

x(x + 2) = 35
x^2 + 2x - 35 = 0
(x + 7)(x - 5) = 0
x = -7, 5

-7 and -5
or
5 and 7

a. f(x) = x^2 - 5x + 6, f'(x) = 2x - 5. f'(x) = 0 at x = (5/2). f''(5/2) = 2, > 0-->f(x) has rel min at the point
P[(5/2) , f(5/2)] = [(5/2) , (5/2 -2)(5/2 - 3)] = [(5/2) , (1/2)(-1/2)] = [(5/2) , - (1/4)]. Vertex is at P & axis
of symmetry is line x = (5/2).
b. Let #'s be a, a+2. Now a(a+2) = 35, ie., a^2 + 2a +1 = 36, ie., (a+1)^2 = 6^2 & a = -1(+/-)6, ie.,
a = -1-6 = -7..(i) or a = -1+6 = 5..(ii). For (i) holding, #'s are -7 & -5. For (ii) holding, #'s are 5 & 7.

Well,

a.
f(x) = x^2 – 5x + 6
= x^2 - 2*5/2 * x + (5/2)^2 - (5/2)^2 + 6
= (x - 5/2)^2 - 25/4 + 24/4
= (x - 5/2)^2 - 1/4
therefore
Vertex : V(5/2, -1/4)
axis of symmetry : x = 5/2

b.
5 and 7

hope it' ll help !!