Chapter Closure
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Exercise 177 Page 116
a
Use the Quadratic Formula to determine the x-intercepts.
b
The function is written in factored form.
c
Use the Quadratic Formula to determine the x-intercepts.
d
Use the Quadratic Formula to determine the x-intercepts.
a
y-intercept: (0,12)
x-intercepts: (- 6,0) and (- 2,0)
Graphing Form: y=(x+4)^2-4
Vertex: (- 4, - 4)
Diagram:
b
y-intercept: (0,- 8)
x-intercepts: (4,0) and (- 2,0)
Graphing Form: y=(x-1)^2-9
Vertex: (1, - 9)
Diagram:
c
y-intercept: (0,- 9)
x-intercepts: (3+sqrt(18),0) and (3-sqrt(18),0)
Vertex: (3,- 18)
Graphing Form: y=(x-3)^2-18
Diagram:
d
y-intercept: (0,1)
x-intercepts: (- 5 - sqrt(21),0 ), and (- 5 + sqrt(21),0 )
Vertex: (- 2.5, - 5.25)
Graphing Form: y=(x+2.5)^2-5.25
Diagram:
a
In any function, the constant tells us the y-coordinate where the function intercepts the y-axis.
y=x^2+8x+ 12 ← constant As we can see, the function intercepts the y-axis at (0,12). To find the x-intercepts, we set y equal to 0 and solve for x with the Quadratic Formula.
y=x^2+8x+12
Substitute
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)
lx_1=- 6 x_2=- 2
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
Substitute
x= - 4
y=( - 4)^2+8( - 4)+12
▼
Evaluate right-hand side
NegBaseToPosBase
(- a)^2=a^2
y=(4)^2+8(- 4)+12
CalcPowProd
Calculate power and product
y=16-32+12
AddSubTerms
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.
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)
Substitute
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)
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.
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
Substitute
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)
CalcPowProd
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.
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
Substitute
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
CalcPowProd
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
Substitute
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
Multiply
Multiply
y=6.25-12.5+1
AddSubTerms
Add and subtract terms
y=-5.25
The vertex is at (-2.5,-5.25). 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-( -2.5))^2+( -5.25) Vertex:& ( -2.5, -5.25) Stretch Factor:& 1 This simplifies to y=(x+2.5)^2-5.25.
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