Let's start by recalling the standard form of a quadratic function.

y = a x^{2} + b x + c

To find the equation of a parabola that includes the given points, we will substitute their coordinates into the above equation and simplify. With the resulting equations, we will write a system of equations. Then, we will solve it to find the coefficients

a,

b,

and

c .

$y = a x^{2} + b x + c$
Point	Substitute	Simplify
$(3, - 6)$	$- 6 = a (3)^{2} + b (3) + c$	$9 a + 3 b + c = - 6$
$(1, - 2)$	$- 2 = a (1)^{2} + b (1) + c$	$a + b + c = - 2$
$(6, 3)$	$3 = a (6)^{2} + b (6) + c$	$36 a + 6 b + c = 3$

We can now write a system of three equations.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 a + 3 b + c = - 6 a + b + c = - 2 36 a + 6 b + c = 3 (I) (II) (III)

Let's solve this system using the Elimination Method. We will start by subtracting Equation (II) from Equation (I) and Equation (III) to eliminate

c .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 a + 3 b + c = - 6 a + b + c = - 2 36 a + 6 b + c = 3

$(I), (III):$ Subtract $(II)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 a + 3 b + c - (a + b + c) = - 6 - (- 2) a + b + c = - 2 36 a + 6 b + c - (a + b + c) = 3 - (- 2)

▼

(I), (III):

Simplify

$(I), (III):$ Distribute $- 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 a + 3 b + c - a - b - c = - 6 - (- 2) a + b + c = - 2 36 a + 6 b + c - a - b - c = 3 - (- 2)

$(I), (III):$ $a - (- b) = a + b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 9 a + 3 b + c - a - b - c = - 6 + 2 a + b + c = - 2 36 a + 6 b + c - a - b - c = 3 + 2

$(I), (III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 8 a + 2 b = - 4 a + b + c = - 2 35 a + 5 b = 5

Now, neither Equation (I) nor Equation (III) includes

c .

These equations form a system in terms of only

a

and

b .

Let's solve this system by using the Elimination Method again. First, we need to simplify the equations a bit.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 8 a + 2 b = - 4 a + b + c = - 2 35 a + 5 b = 5 (I) (II) (III)

▼

(I), (III):

Simplify

FactorOut

$(I):$ Factor out $2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 (4 a + b) = - 4 a + b + c = - 2 35 a + 5 b = 5

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 35 a + 5 b = 5

FactorOut

$(III):$ Factor out $5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 5 (7 a + b) = 5

DivEqn

$(III):$ $LHS / 5 = RHS / 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 7 a + b = 1

SysEqnSub

$(III):$ Subtract $(I)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 7 a + b - (4 a + b) = 1 - (- 2)

▼

(III):

Solve for

a

Distr

$(III):$ Distribute $- 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 7 a + b - 4 a - b = 1 - (- 2)

SubNeg

$(III):$ $a - (- b) = a + b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 7 a + b - 4 a - b = 1 + 2

AddSubTerms

$(III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 3 a = 3

DivEqn

$(III):$ $LHS / 3 = RHS / 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 a = 1

We found our first value, allowing us to write a partial equation.

y = 1 x^{2} + b x + c ⇕ y = x^{2} + b x + c

Let's substitute

1

for

a

in Equation (I) to find the value of

b .

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 a + b = - 2 a + b + c = - 2 a = 1 (I) (II) (III)

Substitute

$(I):$ $a = 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 (1) + b = - 2 a + b + c = - 2 a = 1

▼

(I):

Solve for

b

IdPropMult

$(I):$ Identity Property of Multiplication

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 + b = - 2 a + b + c = - 2 a = 1

SubEqn

$(I):$ $LHS - 4 = RHS - 4$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 6 a + b + c = - 2 a = 1

With our second value, we can continue writing our equation.

y = x^{2} + (- 6) x + c ⇕ y = x^{2} - 6 x + c

Finally, to find the value of

c,

we will substitute

a = 1

and

b = - 6

into Equation (II).

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 6 a + b + c = - 2 a = 1 (I) (II) (III)

SubstituteII

$(II):$ $a = 1$ , $b = - 6$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 6 1 + (- 6) + c = - 2 a = 1

▼

(II):

Solve for

c

AddNeg

$(II):$ $a + (- b) = a - b$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 6 - 5 + c = - 2 a = 1

AddEqn

$(II):$ $LHS + 5 = RHS + 5$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 6 c = 3 a = 1

Now that we have all three values, we can complete the equation of the parabola that passes through the given points.

y = x^{2} - 6 x + 3

To help visualize this situation, we can plot the given points and sketch the curve.

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