Find an equation in standard form of the parabola passing through (3,-6),(1,-2),(6,3) Help me please

y = x^2 - 6x + 3

Step-by-step explanation:

let the equation of the parabola (in standard form) be y = ax^2 + bx + c

sub (3,-6), (1,-2) and (6,3):

-6 = a(3)^2 + b(3) + c

-6 = 9a + 3b + c

c = -6 - 9a - 3b --(1)

-2 = a(1)^2 + b(1) + c

-2 = a + b + c --(2)

3 = a(6)^2 + b(6) + c

3 = 36a + 6b + c --(3)

sub (1) into (2):

-2 = a + b - 6 - 9a - 3b

b = -(4a + 2) --(4)

sub (1) and (4) into (3):

3 = 36a + 6(-4a-2) - 6 - 9a - 3(-4a-2)

3 = 36a -24a - 12 - 6 - 9a + 12a + 6

15a = 15

a = 1

sub a = 1 into (4):

b = -(4(1) + 2)

b = -6

sub a = 1 and b = -6 into (1):

c = -6 - 9(1) - 3(-6)

c = 3

therefore, equation of parabola is y = x^2 - 6x + 3

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example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.