## Pages

### Graph quadratic y = (x - 5)^2

Comparing with vertex form y = a(x - h)^2 + k,

• a = 1
• h = 5
• k = 0
Therefore vertex coordinates are (5, 0)
Axis of symmetry is x = 5

#### x-intercepts

(x - 5)^2 = 0
x - 5 = 0
x = 5
x-intercept is (5, 0). Since the x-intercept coincides with the vertex, find more points on the graph

#### Points

• x = 4, y = (4 - 5)^2 = 1
• x = 6, y = (6 - 5)^2 = 1
• x = 3, y = (3 - 5)^2 = 4
• x = 7, y = (7 - 5)^2 = 4
• x = 2, y = (2 - 5)^2 = 9
• x = 8, y = (8 - 5)^2 = 9
Points are (4, 1), (6, 1), (3, 4), (7, 4), (2, 9) and (8, 9). Plot the vertex and points and join them with a free hand curve. The following graph shows these points along with the parabola.
 Graph y = (x - 5)^2

### Graph quadratic y = 2(x - 3)^2 + 4

Comparing with vertex form y = a(x - h)^2 + k,

• a = 2
• h = 3
• k = 4
Vertex coordinates are (h, k) that is (3, 4)
Axis of symmetry is x = h that is x = 3

#### x-intercepts

2(x - 3)^2 + 4 = 0
2(x - 3)^2 = -4
(x - 3)^2 = -2
x - 3 = sqrt(-2)
Since square root of negative numbers is not a real number, so the equation does not have x intercepts. So you need to get two or more other points on the graph of the parabola:

#### Points

• x = 1, y = 2(1 - 3)^2 + 4 = 12
• x = 5, 'y = 2(5 - 3)^2 + 4 = 12'
You can get more points if you like.
Plot the vertex and points and join them with a free hand curve:
 Graph of y = 2(x - 3)^2 + 4

### Graph quadratic y = -3(x + 2)^2 + 1

Comparing with vertex form y = a(x - h)^2 + k,
• a = -3
• h = -2
• k = 1
Vertex is (-2, 1)
Axis of symmetry is x = -2

#### x-intercepts

-3(x + 2)^2 + 1 = 0
-3(x + 2)^2 = -1
(x + 2)^2 = -1/-3
x + 2 = ± sqrt(1/3) = ± 0.5774
x = -2 + 0.58 or x = -2 - 0.58
x = -1.42 or x = -2.58
x-intercepts are (-1.42, 0) and (-2.58, 0)

#### Points

• x = -3, y = -3(-3 + 2)^2 + 1 = -2
• x = -1, y = -3(-1 + 2)^2 + 1 = -2
Points are (-3, -2) and (-1, -2)
 Graph of y = -3(x + 2)^2 + 1