Exercise 177 Page &nbsp; 116

y= 0

0=x^2+8x+12

UseQuadForm

Use the Quadratic Formula: a = 1, b= 8, c= 12

x=- 8±sqrt(8^2-4( 1)( 12))/2( 1)

▼

Evaluate right-hand side

Calculate power and product

x=- 8± sqrt(64-48)/2

SubTerm

Subtract term

x=- 8± sqrt(16)/2

CalcRoot

Calculate root

x=- 8± 4/2

CalcQuot

Calculate quotient

x=- 4± 2

StateSol

State solutions

lx=-4-2 x=-4+2

CalcQuot

Calculate quotient

lx_1=- 6 x_2=- 2

All parabolas are symmetric about their vertex. What this means is if two points have the same y-coordinate, such as the x-intercepts, they are equidistant from the parabola's line of symmetry. Therefore, we can find the line of symmetry by averaging the x-intercepts.

To find the vertex y-coordinate, we substitute its x-coordinate into the function and evaluate the right-hand side.

y=x^2+8x+12

x= - 4

y=( - 4)^2+8( - 4)+12

▼

Evaluate right-hand side

NegBaseToPosBase

(- a)^2=a^2

y=(4)^2+8(- 4)+12

Calculate power and product

y=16-32+12

Add and subtract terms

y=- 4

To write the function in graphing form, we need to know its vertex and stretch factor. Graphing Form:& y= a(x- h)^2+ k Vertex:& ( h, k) Stretch Factor:& a We have already calculated the vertex. From the equation, we also know the stretch factor since this is the same thing as the squared variable's coefficient. Let's substitute this into the graphing form. Function:& y= 1(x-( -4))^2+( -4) Vertex:& ( -4, -4) Stretch Factor:& 1 This simplifies to y=(x+4)^2-4.

b Let's start by finding the function's y-intercept. We can find this by substituting x=0 into the equation and simplifying the right-hand side.

y=(x-4)(x+2)

x= 0

y=( 0-4)( 0+2)

▼

Evaluate right-hand side

Add and subtract terms

y=(-4)(2)

y=-8

The y-intercept is at (0,-8). Examining the equation, we notice that it's written in factored form. This means we can find its x-intercepts by setting y equal to 0 and solving for x with the Zero Product Property.

y=(x-4)(x+2)

y= 0

0=(x-4)(x+2)

RearrangeEqn

Rearrange equation

(x-4)(x+2)=0

ZeroProdProp

Use the Zero Product Property

lcx-4=0 & (I) x+2=0 & (II)

▼

Solve for x

AddEqn

(I): LHS+4=RHS+4

lx=4 x+2=0

SubEqn

(II): LHS-2=RHS-2

lx_1=4 x_2=- 2

Our function intersects the x-axis twice, at (4,0) and (- 2,0). Like in Part A, we can find the parabola's line of symmetry by averaging the x-intercepts.

The function has its vertex at x=1. By substituting this x-coordinate into the function, we can determine the y-coordinate.

y=(x-4)(x+2)

x= 1

y=( 1-4)( 1+2)

▼

Evaluate right-hand side

Add and subtract terms

y=(-3)(3)

y=-9

The vertex is (1,- 9). To write the function in graphing form, we need to know its vertex and stretch factor. Graphing Form:& y= a(x- h)^2+ k Vertex:& ( h, k) Stretch Factor:& a We have already calculated the vertex. From the equation, we also know the stretch factor since this is the same thing as the squared variable's coefficient. Let's substitute this into the graphing form. Function:& y= 1(x- 1)^2+( -9) Vertex:& ( 1, -9) Stretch Factor:& 1 This simplifies to y=(x-1)^2-9.

c Like in Part A, we have been given the function written in standard form.

y=x^2-6x - 9 ← constant As we can see, the y-intercept is at (0,- 9). Let's solve for the x-intercepts by using the Quadratic Formula.

y=x^2-6x-9

y= 0

0=x^2-6x-9

UseQuadForm

Use the Quadratic Formula: a = 1, b= - 6, c= - 9

x=- ( - 6)±sqrt(( - 6)^2-4( 1)( - 9))/2( 1)

▼

Evaluate right-hand side

NegNeg

- (- a)=a

x=6± sqrt((- 6)^2-4(1)(- 9))/2(1)

Calculate power and product

x=6± sqrt(36+36)/2

SumToProdTwoTerms

a+a=2a

x=6± sqrt((36)2)/2

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=6± sqrt(36)sqrt(2)/2

CalcRoot

Calculate root

x=6± 6sqrt(2)/2

CalcQuot

Calculate quotient

x=3± 3sqrt(2)

StateSol

State solutions

lx_1=3-3sqrt(2) x_2=3+3sqrt(2)

Like in previous parts, we will find the vertex x-coordinate by averaging the x-intercepts.

The graph has its vertex at x=3. By substituting this x-coordinate into the function, we can determine the vertex y-coordinate.

y=x^2-6x-9

x= 3

y=( 3)^2-6( 3)-9

▼

Evaluate right-hand side

CalcPow

Calculate power

y=9-6(3)-9

y=9-18-9

SubTerm

Subtract term

y=-18

The vertex is (3,- 18). By substituting the vertex and stretch factor in the graphing form of a parabola, we can write it in graphing form. Function:& y= 1(x- 3)^2+( -18) Vertex:& ( 3, -18) Stretch Factor:& 1 This simplifies to y=(x-3)^2-18.

d Like in Part A and C, we can directly find the y-intercept by identifying the function's constant.

y=x^2+5x+ 1 ← constant As we can see, the y-intercept is at (0,1). Let's solve for the x-intercepts by using the Quadratic Formula.

y=x^2+5x+1

y= 0

0=x^2+5x+1

UseQuadForm

Use the Quadratic Formula: a = 1, b= 5, c= 1

x=- 5±sqrt(5^2-4( 1)( 1))/2( 1)

▼

Evaluate right-hand side

Calculate power and product

x=-5± sqrt(25-4)/2

SubTerm

Subtract term

x=-5± sqrt(21)/2

StateSol

State solutions

lx_1= -5-sqrt(21)2 x_2= -5+sqrt(21)2

Like in previous parts, we will find the vertex x-coordinate by averaging the x-intercepts.

The graph has its vertex at x=- 2.5. By substituting this x-coordinate into the function, we can determine the vertex y-coordinate.

y=x^2+5x+1

x= -2.5

y=( -2.5)^2+5( -2.5)+1

▼

Evaluate right-hand side

NegBaseToPosBase

(- a)^2=a^2

y=(2.5)^2+5(-2.5)+1

CalcPow

Calculate power

y=6.25+5(-2.5)+1

MultPosNeg

a(- b)=- a * b

y=6.25-5(2.5)+1

y=6.25-12.5+1