Embed Size (px)
Citation preview
Graphing Parabolas
Using the Vertex
Axis of Symmetry
& y-Intercept
By: Jeffrey Bivin
Lake Zurich High School
Last Updated: October 15, 2007
Graphing Parabolas
y = x2 + 4x - 7
• With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry.
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
Jeff Bivin -- LZHS
Graph the following parabola
y = x2 + 4x - 7
axis of symmetry:
vertex:
2x
)11,2(
)7,0( y-intercept:
Jeff Bivin -- LZHS
Graph the following parabola
y = 2x2 + 10x + 4
axis of symmetry:
vertex:
25x
),( 217
25
)4,0(y-intercept:
25x
),( 217
25
)4,0(
Jeff Bivin -- LZHS
Graph the following parabola
y = -3x2 + 5x + 9
axis of symmetry:
vertex:
65x
),( 12133
65
)9,0(y-intercept:
65x
),( 12133
65
)9,0(
Jeff Bivin -- LZHS
Graphing Parabolas
y = x2 + 4x - 7
• Now look at the coefficients of the equation and the value of the axis of symmetry – especially a and b
• y = ax2 + bx + c
y = 2x2 + 10x + 4
y = -3x2 + 5x + 9
2x
25x
65x
Jeff Bivin -- LZHS
Graphing Parabolas
y = ax2 + bx + c
Vertex:
Axis of symmetry: abx 2
)(, 22 ab
ab f
Jeff Bivin -- LZHS
Graph the following parabola
y = x2 + 4x - 7
axis of symmetry:
vertex:
224
)1(24
2 abx
)11,2(
)7,0( y-intercept:
117)2(4)2()2( 2 f
Jeff Bivin -- LZHS
Graph the following parabola
y = 2x2 + 10x + 4
axis of symmetry:
vertex:
25
410
)2(210
2 a
bx
),( 217
25
)4,0(y-intercept:
217
252
25
25 4)(10)(2)( f
25x
Jeff Bivin -- LZHS
),( 217
25
Graph the following parabola
y = -3x2 + 5x + 9
axis of symmetry:
vertex:
65
65
)3(25
2
abx
),( 12133
65
)9,0(y-intercept:
12133
652
65
65 9)(5)(3)( f
65x
),( 12133
65
)9,0(
Why did this parabola open downward instead of upward
as did the previous two?
Jeff Bivin -- LZHS
Graph the following parabolay = x2 + 6x - 8
Axis of symmetry:
Vertex:
3)1(26
2 abx
178)3(6)3()3( 2 f
)17,3(
)8,0( y-intercept:Jeff Bivin -- LZHS
Graph the following parabolay = -2x2 + 7x + 12
Axis of symmetry:
Vertex:
47
47
)2(27
2
abx
8145
472
47
47 12)(7)(2)( f
),( 8145
47
)12,0(y-intercept:
),( 8145
47
47x
Jeff Bivin -- LZHS
Graphing Parabolas
In Vertex Form
Jeff Bivin -- LZHS
Graphing Parabolas
y = x2
• With your graphing calculator, graph each of the following quadratic equations and identify the vertex and axis of symmetry.
y = (x - 5)2 - 4
y = -3(x + 2)2 + 5
y = ⅜(x - 3)2 + 1
Jeff Bivin -- LZHS
0x)0,0(
5x)4,5(
2x)5,2(
3x)1,3(
vertex axis of sym.
Graph the following parabola
y = (x - 5)2 - 4
axis of symmetry:
vertex:
5x
)4,5(
)21,0(y-intercept:
Jeff Bivin -- LZHS
05 x
Graph the following parabola
y = -3(x + 2)2 + 5
axis of symmetry:
vertex:
2x
)5,2(
)7,0( y-intercept:
Jeff Bivin -- LZHS
02 x
Graph the following parabola
y = ⅜•(x - 3)2 - 1
axis of symmetry:
vertex:
3x
)1,3(
),0( 819y-intercept:
Jeff Bivin -- LZHS
03 x
),0( 819
Graphing Parabolas
In Intercept Form
Jeff Bivin -- LZHS
Graph the following parabola
y = (x – 4)(x + 2)
x-intercepts:
vertex:
)8,0( y-intercept:Jeff Bivin -- LZHS
04x 02x
)0,4( )0,2(
axis of symmetry:
224 x
1x)9,1(
9)3)(3()21)(41( y
8)2)(4()20)(40( y
Graph the following parabola
y = (x - 1)(x - 9)
x-intercepts:
vertex:
)9,0(y-intercept:Jeff Bivin -- LZHS
01x 09x
)0,1( )0,9(
axis of symmetry:
291 x
5x)16,5(
16)4)(4()95)(15( y
9)9)(1()90)(10( y
Graph the following parabola
y = -2(x + 1)(x - 5)
x-intercepts:
vertex:
)10,0(y-intercept:Jeff Bivin -- LZHS
01x 05x
)0,1( )0,5(
axis of symmetry:
251 x
2x)18,2(
18)3)(3(2)52)(12(2 y
10)5)(1(2)50)(10(2 y
Convert to standard form
y = -2(x + 1)(x - 5)
Jeff Bivin -- LZHS
y = -2(x2 – 5x + 1x – 5)
y = -2(x2 – 4x – 5)
y = -2x2 + 8x + 10
Now graph from standard form.y = -2x2 + 8x + 10
Axis of symmetry:
Vertex:
248
)2(28
2
abx
1810)2(8)2(2)2( 2 f
)18,2(
)10,0(y-intercept:Jeff Bivin -- LZHS
A taxi service operates between two airports transporting 200 passengers a day. The charge is $15.00. The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income?
Jeff Bivin -- LZHS
Income = Price ● Quantity
f(x) = ( 15 + 2x ) ( 200 – 10x )
Define the variable
x = number of $2 price increases 15 + 2x = 0 200 – 10x = 0
25.3781)25.6(10200)25.6(215)25.6( f
2x = -15Vertex is:
25.3781,25.6
So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50
f(x) = income
200 = 10x
215x x20
25.6: 425
2240
215
xsymmetryofaxis
Maximumincome:
VERTEX
27.50 137.50
A taxi service operates between two airports transporting 200 passengers a day. The charge is $15.00. The owner estimates that 10 passengers will be lost for each $2 increase in the fare. What charge would be most profitable for the service? What is the maximum income?
Jeff Bivin -- LZHS
Income = Price ● Quantity
f(x) = ( 15 + 2x ) ( 200 – 10x )
Define the variable
x = number of $2 price increases f(x) = 3000 – 150x + 400x – 20x2
f(x) = – 20x2 + 250x + 3000
VERTEX
abx 2
)20(2250
x25.6x
f(6.25) = – 20(6.25)2 + 250(6.25) + 3000
f(6.25) = 3781.25 Vertex is:
25.3781,25.6
So, price = (15 + 2x) = (15 + 2(6.25)) = 15 + 12.5 = $27.50
f(x) = income
Maximum income = f(x) = $3781.25
