−
b
a2
• The standard form of a quadratic equation is y = ax2
+ bx + c.
• The graph of a quadratic equation is a parabola....
To find the x coordinate of the vertex, use the equation
Then substitute the value of x back into the equation of the
para...
Choose two values of x that are to the right or left of the x-
coordinate of the vertex.
Substitute those values in the ...
x
y
Plot the vertex and the points from
your table of values: (2,3), (1,2),
(-1,-6).
Since a parabola is symmetric and
has...
Find the vertex of the following quadratic equations. Make a
table of values and graph the parabola.
1 42
. y x x= −
2 2 3...
y x x= −2
4
a b= = −1 4
x
b
a
x
x
= −
= −
−
=
2
4
2 1
2
( )
y x x
y
y
y
= −
= −
= −
= −
2
2
4
2 4 2
4 8
4
( )
The vertex i...
y x= − +2 32
a b= − =2 0
x
b
a
x
x
= −
= −
−
=
2
0
2 2
0
( )
y x
y
y
= − +
= − +
=
2 3
2 0 3
3
2
2
( )
The vertex is at (0...
y x x= − +2
6 4
a b= = −1 6
x
b
a
x
x
= −
= −
−
=
2
6
2 1
3
( )
y x x
y
y
= − +
= − +
= −
2
2
6 4
3 6 3 4
5
( )
The vertex...
y x x= − +2
6 4
a b= = −1 6
x
b
a
x
x
= −
= −
−
=
2
6
2 1
3
( )
y x x
y
y
= − +
= − +
= −
2
2
6 4
3 6 3 4
5
( )
The vertex...
Graphing quadratic-equations-4818
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Graphing quadratic-equations-4818

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Find vertex, roots and sketch parabolas

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Graphing quadratic-equations-4818

  1. 1. − b a2 • The standard form of a quadratic equation is y = ax2 + bx + c. • The graph of a quadratic equation is a parabola. • When a is positive, the graph opens up. • When a is negative, the graph opens down. • Every parabola has a vertex. For graphs opening up, the vertex is a minimum (low point). For graphs opening down, the vertex is a maximum (high point). • The x-coordinate of the vertex is equal to .
  2. 2. To find the x coordinate of the vertex, use the equation Then substitute the value of x back into the equation of the parabola and solve for y. x b a = − 2 You are given the equation y=-x2 + 4x –1. Find the coordinates of the vertex. x b a = − 2 x = 2 −= =a b1 4, ( − x = − 4 2 1)( ) 2 y x x= − + −4 1 y = − + −4 8 1 y = 3 y = − + −2 2 4 2 1( ) ( ) The coordinates of the vertex are (2,3)
  3. 3. Choose two values of x that are to the right or left of the x- coordinate of the vertex. Substitute those values in the equation and solve for y. Graph the points. Since a parabola is symmetric, you can plot those same y-values on the other side of the parabola, the same distance away from the vertex. x y = -x2 + 4x -1 y 1 -1 y = -(1)2 + 4(1) - 1 y = -1 +4 - 1 2 y = -(-1)2 + 4(-1) -1 y = -1 -4 -1 -6
  4. 4. x y Plot the vertex and the points from your table of values: (2,3), (1,2), (-1,-6). Since a parabola is symmetric and has a vertical line of symmetry through the vertex of the parabola, points which are the same distance to the right of the vertex will also have the same y-coordinate as those to the left of the vertex. Therefore, we can plot points at (3,2) and (5,-6). Sketch in the parabola.
  5. 5. Find the vertex of the following quadratic equations. Make a table of values and graph the parabola. 1 42 . y x x= − 2 2 32 . y x= − + 3 6 42 . y = x x− +
  6. 6. y x x= −2 4 a b= = −1 4 x b a x x = − = − − = 2 4 2 1 2 ( ) y x x y y y = − = − = − = − 2 2 4 2 4 2 4 8 4 ( ) The vertex is at (2,-4) x y = x 2 - 4 x y 1 y = 1 2 - 4 ( 1 ) - 3 0 y = 0 2 - 4 ( 0 ) 0 x y
  7. 7. y x= − +2 32 a b= − =2 0 x b a x x = − = − − = 2 0 2 2 0 ( ) y x y y = − + = − + = 2 3 2 0 3 3 2 2 ( ) The vertex is at (0,3) x y = - 2 x 2 + 3 y - 1 y = - 2 ( - 1 ) 2 + 3 1 - 2 y = - 2 ( - 2 ) 2 + 3 - 5 x y
  8. 8. y x x= − +2 6 4 a b= = −1 6 x b a x x = − = − − = 2 6 2 1 3 ( ) y x x y y = − + = − + = − 2 2 6 4 3 6 3 4 5 ( ) The vertex is at (3,-5) x y = x 2 - 6 x + 4 y 1 y = 1 2 - 6 ( 1 ) + 4 - 1 0 y = 0 2 - 6 ( 0 ) + 4 4 x y
  9. 9. y x x= − +2 6 4 a b= = −1 6 x b a x x = − = − − = 2 6 2 1 3 ( ) y x x y y = − + = − + = − 2 2 6 4 3 6 3 4 5 ( ) The vertex is at (3,-5) x y = x 2 - 6 x + 4 y 1 y = 1 2 - 6 ( 1 ) + 4 - 1 0 y = 0 2 - 6 ( 0 ) + 4 4 x y

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