Exercise 4 Page 493

The relationship between time t and the height h of any ball tossed into the air is always modeled by a quadratic equation. h=at^2+bt+c, a≠ 0 In order to determine which equation models the relationship for the given ball, we have to find the values of a, b, and c. To do so we will write a system of equations using three of the given points: (0,4), (0.25,10.5), and (0.5,15). Let's start with (0,4). We just wrote and solved our first equation. Now let's write a second equation using the point (0.25,10.5).

h=at^2+bt+c

SubstituteII

t= 0.25, h= 10.5

10.5=a( 0.25)^2+b( 0.25)+c

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Simplify

CalcPow

Calculate power

10.5=a(0.0625)+b(0.25)+c

Multiply

Multiply

10.5=0.0625a+0.25b+c

RearrangeEqn

Rearrange equation

0.0625a+0.25b+c=10.5

To find our third and last equation, we will use (0.5,15).

h=at^2+bt+c

SubstituteII

t= 0.5, h= 15

15=a( 0.5)^2+b( 0.5)+c

▼

Simplify

CalcPow

Calculate power

15=a(0.25)+b(0.5)+c

Multiply

Multiply

15=0.25a+0.5b+c

RearrangeEqn

Rearrange equation

0.25a+0.5b+c=15

We now have a system of three equations. c=4 & (I) 0.0625a+0.25b+c=10.5 & (II) 0.25a+0.5b+c=15 & (III) Since Equation (I) is already solved for c, we will substitute 4 for c into Equation (II) and Equation (III).

c=4 0.0625a+0.25b+c=10.5 0.25a+0.5b+c=15

(II), (III): c= 4

c=4 0.0625a+0.25b+ 4=10.5 0.25a+0.5b+ 4=15

(II), (III): LHS-4=RHS-4

c=4 0.0625a+0.25b=6.5 0.25a+0.5b=11

Next we will use the Elimination Method. We will start by multiplying both sides of Equation (III) by 0.5. Our goal is to eliminate the b-variable from Equation (II) and solve it for a.

c=4 0.0625a+0.25b=6.5 0.25a+0.5b=11

MultEqn

(III): LHS * 0.5=RHS* 0.5

c=4 0.0625a+0.25b=6.5 0.125a+0.25b=5.5

SysEqnSub

(II): Subtract (III)

c=4 0.0625a+0.25b-( 0.125a+0.25b)=6.5- 5.5 0.125a+0.25b=5.5

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(II):Solve for a

Distr

(II): Distribute - 1

c=4 0.0625a+0.25b-0.125a-0.25b=6.5-5.5 0.125a+0.25b=5.5

SubTerms

(II): Subtract terms

c=4 - 0.0625a=1 0.125a+0.25b=5.5

DivEqn

(II): .LHS /(- 0.0625).=.RHS /(- 0.0625).

c=4 a=1/- 0.0625 0.125a+0.25b=5.5

UseCalc

(II): Use a calculator

c=4 a=- 16 0.125a+0.25b=5.5

Next, we will substitute - 16 for a in Equation (III) to find the value of b.

c=4 a=- 16 0.125a+0.25b=5.5

Substitute

(III): a= - 16

c=4 a=- 16 0.125( - 16)+0.25b=5.5

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(III):Solve for b

MultPosNeg

(III): a(- b)=- a * b

c=4 a=- 16 - 2+0.25b=5.5

AddEqn

(III): LHS+2=RHS+2

c=4 a=- 16 0.25b=7.5

DivEqn

(III): .LHS /0.25.=.RHS /0.25.

c=4 a=- 16 b=30

Since we have all three values, we can complete the quadratic equation that models the relationship between time t and the height h of the ball tossed into the air. h=- 16t^2+30t+4 This corresponds to option G.