Exercise 3 Page 212

Let's start by recalling the standard form of a quadratic function. y=ax^2+bx+c To find the equation of a parabola that includes the given points, we will substitute the coordinates of three of them 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. Let's consider the table.

y=ax^2+bx+c
Point	Substitute	Simplify
( 0, - 13.5)	- 13.5=a( 0)^2+b( 0)+c	c= - 13.5
( - 1, - 25.5)	- 25.5=a( - 1)^2+b( - 1)+c	a-b+c=- 25.5
( 1, - 5.5)	- 5.5=a( 1)^2+b( 1)+c	a+b+c=- 5.5

From the first equation, we already know that c= - 13.5. With this information, we can write a partial equation of the parabola. y=ax^2+bx+( - 13.5) ⇕ y=ax^2+bx-13.5 We can now write a system of two equations. a-b+c=- 25.5 a+b+c=- 5.5 ⇓ a-b+( - 13.5)=- 25.5 & (I) a+b+( - 13.5)=- 5.5 & (II) Let's solve this system by using the Elimination Method. Since b has opposite coefficients in both equations, we will start by adding Equation (II) to Equation (I).

a-b+(- 13.5)=- 25.5 a+b+(- 13.5)=- 5.5

▼

(I), (II): Simplify

(I), (II): a+(- b)=a-b

a-b-13.5=- 25.5 a+b-13.5=- 5.5

(I), (II): LHS+13.5=RHS+13.5

a-b=- 21 a+b=8

SysEqnAdd

(I): Add (II)

a-b+ a+b=- 12+ 8 a+b=8

▼

(I): Solve for a

AddTerms

(I): Add terms

2a=- 4 a+b=8

DivEqn

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

a= - 2 a+b=8

We found that a= - 2. We can add this information to our equation. y= - 2x^2+bx-13.5 Let's now substitute - 2 for a in Equation (II).

a=- 2 & (I) a+b=8 & (II)

Substitute

(II): a= - 2

a=- 2 - 2+b=8

AddEqn

(II): LHS+2=RHS+2

a=- 2 b= 10

Now that we have all three values, we can complete the equation of the parabola that passes through the given points. y=- 2x^2+ 10x-13.5 To help visualize this situation, we can plot the given points and sketch the curve.