Exercise 33 Page 967

We are given that the quadratic function

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

passes through the following three points.

(0, 0), (1, 2), (3, - 6)

We will first find the coefficients and the constant of our equation. To do so, since those three points satisfy our equation we will substitute them into the equation one at a time. Let's begin with

(0, 0) .

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

SubstituteII

$x = 0$ , $y = 0$

0 = a \cdot 0^{2} + b \cdot 0 + c

▼

Solve for

c

CalcPow

Calculate power

0 = a \cdot 0 + b \cdot 0 + c

ZeroPropMult

Zero Property of Multiplication

0 = 0 + 0 + c

IdPropAdd

Identity Property of Addition

0 = c

RearrangeEqn

Rearrange equation

c = 0

Great! Now that we have

c = 0,

we will next substitute

(1, 2)

into the equation and solve it one more time.

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

SubstituteValues

Substitute values

2 = a \cdot 1^{2} + b \cdot 1 + 0

▼

CalcPow

Calculate power

2 = a \cdot 1 + b \cdot 1 + 0

IdPropMult

Identity Property of Multiplication

2 = a + b + 0

IdPropAdd

Identity Property of Addition

2 = a + b

RearrangeEqn

Rearrange equation

a + b = 2

Last, we will substitute

(3, - 6)

into the equation.

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

SubstituteValues

Substitute values

- 6 = a \cdot 3^{2} + b \cdot 3 + 0

▼

CalcPow

Calculate power

- 6 = a \cdot 9 + b \cdot 3 + 0

CommutativePropMult

Commutative Property of Multiplication

- 6 = 9 a + 3 b + 0

IdPropAdd

Identity Property of Addition

- 6 = 9 a + 3 b

RearrangeEqn

Rearrange equation

9 a + 3 b = - 6

DivEqn

$LHS / 3 = RHS / 3$

\frac{9 a + 3 b}{3} = \frac{- 6}{3}

WriteSumFrac

Write as a sum of fractions

\frac{9 a}{3} + \frac{3 b}{3} = \frac{- 6}{3}

CalcQuot

Calculate quotient

3 a + b = - 2

Since we got two equations with the same variables, we will now solve the following system.

{a + b = 2 3 a + b = - 2

To do so we can use the Elimination Method by subtracting the first equation from the second one.

{a + b = 2 3 a + b = - 2 (I) (II)

SysEqnSub

$(II):$ Subtract $I$

{a + b = 2 3 a + b - (a + b) = - 2 - 2 (I) (II)

Distr

$(II):$ Distribute $- 1$

{a + b = 2 3 a + b - a - b = - 2 - 2 (I) (II)

AddSubTerms

$(II):$ Add and subtract terms

{a + b = 2 2 a = - 4 (I) (II)

DivEqn

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

{a + b = 2 a = - 2 (I) (II)

Substitute

a = - 2

into the first equation to solve for

b .

{a + b = 2 a = - 2 (I) (II)

Substitute

$(I):$ $a = - 2$

{- 2 + b = 2 a = - 2 (I) (II)

AddEqn

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

{b = 4 a = - 2 (I) (II)

Having all the coefficients, we can complete our equation. Let's see it!

y = - 2 x^{2} + 4 x + 0

Now we will find the axis of symmetry of the given parabola. To do so, remember the formula for the axis of symmetry.

x = - \frac{b}{2 a}

Since we have found that

a = - 2

and

b = 4,

we can substitute those values and solve for

x .

x = - \frac{b}{2 a}

SubstituteII

$a = - 2$ , $b = 4$

x = - \frac{4}{2 \cdot ( - 2 )}

▼

Solve for

x

MultPosNeg

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

x = - \frac{4}{- 4}

MoveNegFracToNum

Put minus sign in numerator

x = \frac{- 4}{- 4}

QuotOne

$\frac{a}{a} = 1$

x = 1

This corresponds to option H.

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