Exercise 127 Page 512

To find all points where the given functions intersect, we will create a system of equations. The solutions of this system are points where the given functions intersect. We will solve this system of equations using the Substitution Method. y=x^2-3x+2 & (I) y=2x+8 & (II) Notice that the y-variable is isolated both equations. Since the expression equal to y in Equation (II) is simpler, let's use that for our initial substitution. Notice that in Equation (I) we have a quadratic equation in terms of only the x-variable. x^2-5x-6=0 ⇕ 1x^2+( - 5)x+( - 6)=0We can substitute a= 1, b= - 5, and c= - 6 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 5)±sqrt(( - 5)^2-4( 1)( - 6))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=5±sqrt((- 5)^2-4(1)(- 6))/2(1)

NegBaseToPosPow

(- a)^2 = a^2

x=5±sqrt(25-4(1)(- 6))/2(1)

IdPropMult

Identity Property of Multiplication

x=5±sqrt(25-4(- 6))/2

MultNegNegOnePar

- a(- b)=a* b

x=5±sqrt(25+24)/2

AddTerms

Add terms

x=5±sqrt(49)/2

CalcRoot

Calculate root

x=5± 7/2

This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.

x=5± 7/2
x_1=5+7/2	x_2=5-7/2
x_1=12/2	x_2=- 2/2
x_1=6	x_2=- 1

Now, consider Equation (II). y=2x+8 We can substitute x=6 and x=- 1 into the above equation to find the values for y. Let's start with x=6.

y=2x+8

Substitute

x= 6

y=2( 6)+8

▼

Solve for y

Multiply

Multiply

y=12+8

AddTerms

Add terms

y=20

We found that y=20 when x=6. One solution of the system, which is a point of intersection of the two given functions, is (6,20). To find the other solution we will substitute - 1 for x in Equation (II) again.

y=2x+8

Substitute

x= - 1

y=2( - 1)+8

▼

Solve for y

MultPosNeg

a(- b)=- a * b

y=- 2+8

AddTerms

Add terms

y=6

We found that y=6 when x=- 1. Therefore, our second solution, which is the other point of intersection of the two given functions, is ( - 1,6).