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, 4), (0, 1), (2, 7)

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, 4) .

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

SubstituteII

$x = - 1$ , $y = 4$

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

▼

Simplify

CalcPow

Calculate power

4 = a (1) + b (- 1) + c

Multiply

4 = a - b + c

RearrangeEqn

Rearrange equation

a - b + c = 4

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

(0, 1) .

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

SubstituteII

$x = 0$ , $y = 1$

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

▼

Simplify

ZeroPropMult

Zero Property of Multiplication

1 = 0 + 0 + c

IdPropAdd

Identity Property of Addition

1 = c

RearrangeEqn

Rearrange equation

c = 1

To find our third and last equation, we will use

(2, 7) .

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

SubstituteII

$x = 2$ , $y = 7$

7 = a (2)^{2} + b (2) + c

▼

Simplify

CalcPow

Calculate power

7 = a (4) + b (2) + c

Multiply

7 = 4 a + 2 b + c

RearrangeEqn

Rearrange equation

4 a + 2 b + c = 7

We now have a system of three equations.

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

Note that we have already found our first value, allowing us to form a partial equation.

y = a x^{2} + b x + 1

Let's solve the system using the Elimination Method. We will start by substituting

1

for

c

in Equation (I) and Equation (III).

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

Substitute

$(I):$ $c = 1$ , $(III):$ $c = 1$

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

SubEqn

$(I):$ $LHS - 1 = RHS - 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a - b = 3 c = 1 4 a + 2 b + 1 = 7

SubEqn

$(III):$ $LHS - 1 = RHS - 1$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a - b = 3 c = 1 4 a + 2 b = 6

Now, let's multiply Equation (I) by

2

and add Equation (I) and Equation (III) to eliminate the

b -

variable.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a - b = 3 c = 1 4 a + 2 b = 6

MultEqn

$(I):$ $LHS \cdot 2 = RHS \cdot 2$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a - 2 b = 6 c = 1 4 a + 2 b = 6

SysEqnAdd

$(III):$ Add $(I)$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a - 2 b = 6 c = 1 4 a + 2 b + 2 a - 2 b = 6 + 6

▼

(III):

Solve for

a

AddSubTerms

$(III):$ Add and subtract terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 2 a - 2 b = 6 c = 1 6 a = 12

DivEqn

$(III):$ $LHS / 6 = RHS / 6$

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

With our second value, we can continue forming the partial equation.

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

Finally, to find the value of

b,

we will substitute

a = 2

into Equation (I).

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

Substitute

$(I):$ $a = 2$

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

▼

(I):

Solve for

b

Multiply

$(I):$ Multiply

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 4 - 2 b = 6 c = 1 a = 2

SubEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ - 2 b = 2 c = 1 a = 2

DivEqn

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ b = - 1 c = 1 a = 2

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

y = 2 x^{2} + (- 1) x + 1 \Leftrightarrow y = 2 x^{2} - x + 1

To help visualize this graph, we have plotted the given points and sketched the curve below.

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