* note * i just noticed i answered for the wrong thing, however i will leave this in case someone wants to learn it
First you would plug in the values of the y-intercept into a generic quadratic equation. A generic quadratic equation is:
y = ax^2 + bx + c
so plugging in the y intercept values of (0, 175) we get
175 = 0x^2 + 0b + c
this shows what c equals since 0 times any number is zero:
c = 175
now the generic vertex form is written as:
y - k = a * (x - h)^2 or
y = a * (x - h)^2 + k
with (h, k ) being the vertex coordinates
I will use 2nd equation and plug in vertex point
y = a * (x - 6)^2 -5
y = a* (x^2 - 12x + 36) - 5
y = ax^2 - 12x + 36a - 5
Since we know that the standard equation is
y = ax^2 +bx + c
we can use what we just found and plug it into the standard equation but only using 1 of its 3 parts at a time (the parts being ax^2, bx, and c). So:
ax^2 = ax^2, this does nothing
bx = -12x, this can be reduced by dividing both sides by x
b = -12
c = 36a - 5, from earlier we know that c = 175 so
175 = 36a -5
180 = 36a
a = 5
now we have all three values and can plug in to get our equation
y = 5x^2 - 12x + 175
