Before we try to write the equation of a specific parabola that passes through some given points, let's make sure we have enough points. We will need at least three.

(1, 0), (- 2, 9), (- 4, 5)

To use the given points, we need to substitute their

(x, y)

coordinate pairs into the standard form of a quadratic equation.

y = a x^{2} + b x + c, a \neq = 0

Doing so will create a system of equations that we can solve for the values of

a,

b,

and

c .

Let's start with

(1, 0) .

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

SubstituteII

$x = 1$ , $y = 0$

0 = a (1)^{2} + b (1) + c

▼

Simplify

BaseOne

$1^{a} = 1$

0 = a (1) + b (1) + c

IdPropMult

Identity Property of Multiplication

0 = a + b + c

RearrangeEqn

Rearrange equation

a + b + c = 0

We just wrote our first equation! Now let's do the same thing for

(- 2, 9) .

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

SubstituteII

$x = - 2$ , $y = 9$

9 = a (- 2)^{2} + b (- 2) + c

▼

Simplify

CalcPow

Calculate power

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

Multiply

9 = 4 a + b (- 2) + c

MultPosNeg

$a (- b) = - a \cdot b$

9 = 4 a - 2 b + c

RearrangeEqn

Rearrange equation

4 a - 2 b + c = 9

To find our third and last equation, we will use

(- 4, 5) .

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

SubstituteII

$x = - 4$ , $y = 5$

5 = a (- 4)^{2} + b (- 4) + c

▼

Simplify

CalcPow

Calculate power

5 = a (16) + b (- 4) + c

Multiply

5 = 16 a + b (- 4) + c

MultPosNeg

$a (- b) = - a \cdot b$

5 = 16 a - 4 b + c

RearrangeEqn

Rearrange equation

16 a - 4 b + c = 5

We now have a system of three equations.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 4 a - 2 b + c = 9 16 a - 4 b + c = 5 (I) (II) (III)

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

c -

variable.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 4 a - 2 b + c = 9 16 a - 4 b + c = 9

SysEqnSub

$(II):$ Subtract $(I)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 4 a - 2 b + c - (a + b + c) = 9 - 0 16 a - 4 b + c = 9

Distr

$(II):$ Distribute $- 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 4 a - 2 b + c - a - b - c = 9 - 0 16 a - 4 b + c = 9

SubTerms

$(II):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 3 a - 3 b = 9 16 a - 4 b + c = 9

Now let's subtract Equation (I) from Equation (III) to eliminate the

c -

variable once more.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 3 a - 3 b = 9 16 a - 4 b + c = 5

SysEqnSub

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 3 a - 3 b = 9 16 a - 4 b + c - (a + b + c) = 5 - 0

Distr

$(III):$ Distribute $- 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 3 a - 3 b = 9 16 a - 4 b + c - a - b - c = 5 - 0

SubTerms

$(III):$ Subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a + b + c = 0 3 a - 3 b = 9 15 a - 5 b = 5

Now, neither Equation (II) nor Equation (III) includes the

c -

variable. These equations form a system with only two variables,

a

and

b .

Let's solve this system by using the Elimination Method again. Since neither variable has the same or opposite coefficients, we will need to multiply one equation by a number first.

Exercise

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