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`


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