Exercise 5 Page 79

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≠ 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). We just wrote our first equation! Now let's do the same thing for (0,1).

y=ax^2+bx+c

SubstituteII

x= 0, y= 1

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

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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=ax^2+bx+c

SubstituteII

x= 2, y= 7

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

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Simplify

CalcPow

Calculate power

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

Multiply

Multiply

7=4a+2b+c

RearrangeEqn

Rearrange equation

4a+2b+c=7

We now have a system of three equations. a-b+c=4 c=1 4a+2b+c=7 Note that we have already found our first value, allowing us to form a partial equation. y=ax^2+bx+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 4a+2b+c=7

Substitute

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

a-b+ 1=4 c=1 4a+2b+ 1=7

SubEqn

(I): LHS-1=RHS-1

a-b=3 c=1 4a+2b+1=7

SubEqn

(III): LHS-1=RHS-1

a-b=3 c=1 4a+2b=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 4a+2b=6

MultEqn

(I): LHS * 2=RHS* 2

2a-2b=6 c=1 4a+2b=6

SysEqnAdd

(III): Add (I)

2a-2b=6 c=1 4a+2b+ 2a- 2b=6+ 6

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(III): Solve for a

AddSubTerms

(III): Add and subtract terms

2a-2b=6 c=1 6a=12

DivEqn

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

2a-2b=6 c=1 a=2

With our second value, we can continue forming the partial equation. y=2x^2+bx+1 Finally, to find the value of b, we will substitute a=2 into Equation (I).

2a-2b=6 c=1 a=2

Substitute

(I): a= 2

2( 2)-2b=6 c=1 a=2

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(I): Solve for b

Multiply

(I): Multiply

4-2b=6 c=1 a=2

SubEqn

(I): LHS-4=RHS-4

- 2b=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=2x^2+(-1)x+1 ⇔ y=2x^2-x+1 To help visualize this graph, we have plotted the given points and sketched the curve below.