Graphing quadratics in vertex form - Solved Examples

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`