Graphing quadratic in standard form - Solved Examples

Graph quadratic function `y = x^2 + 2x`

This quadratic equation is in standard form `y = ax^2 + bx + c` without the number `c`. In order to graph it, you need to get the vertex coordinates, axis of symmetry, x-intercepts and points on it:
Comparing the equation with `y = ax^2 + bx + c`,

• `a = 1`
• `b = 2`
• `c = 0`

Vertex coordinates

Let the vertex coordinates be `(h, k)`. Applying formula `h = -b/(2a)`,

`h = -2/(2*1) = -1`

Substitute x = h in the equation to get the value of k,

`k = (-1)^2 + 2(-1) = -1`

Vertex coordinates are `(-1, -1)`

Axis of symmetry

Axis of symmetry is the x-coordinate of the vertex `x = -1`

x-intercepts

Equate the function to zero and solve for x

`x^2 + 2x = 0`
`x(x + 2) = 0`
`x = 0, x + 2 = 0`
`x = 0, x = -2`

x-intercepts are (0, 0) and (-2, 0)

Points

In general you don't need to get the points if you got the x-intercepts and if the x-intercept does not coincide with the vertex. Nevertheless the necessary work required to get the points is shown here:

• x = -3, `y = (-3)^2 + 2(-3) = 3` ... point is (-3, 3)
• x = 1, `y = (1)^2 + 2(1) = 3` ... point is (1, 3)

Plot the vertex, x-intercepts, points and join them with a free hand curve

Graph of `y = x^2 + 2x`

Graph quadratic function `y = x^2 - 1`

This is pure quadratic equation and it can be easily rewritten in the standard, vertex or intercepts form.
Comparing it with standard form `y = ax^2 + bx + c`,

• `a = 1`
• `b = 0`
• `c = -1`

Vertex coordinates

Let the vertex coordinates be (h, k). Applying formula `h = -b/(2a)`,

`h = -0/(2*1) = 0`

Substitute x = h in the equation to get the value of k,

`k = (0)^2 - 1 = -1` 

Vertex is (0, -1)

Axis of symmetry

Axis of symmetry is the x-coordinate of the vertex. Hence the axis of symmetry is `x = 0`

x-intercepts

Equate the quadratic equation to zero and solve for x

`x^2 - 1 = 0`
`x^2 - 1^2 = 0`
`(x + 1)(x - 1) = 0`
`x + 1 = 0 or x - 1 = 0`
`x = -1 or x = 1`

x-intercepts are (-1, 0) and (1, 0)

Points

In general it is not necessary to get the points if the parabolas has x-intercepts but you can get more points to make the graph more accurate

• x = -2, `y = (-2)^2 - 1 = 3`... point is (-2, 3)
• x = 2, `y = (2)^2 - 1 = 3` ... point is(2, 3)

Plot the vertex, x-intercepts and points and join them with a free hand curve

Graph quadratic function `y = x^2 + x - 2`

Comparing with `y = ax^2 + bx + c`,

• a = 1
• b = 1
• c = -2

Vertex coordinates

`h = -b/(2a) = -1/(2*1) = -1/2`

`k = (-1/2)^2 + -1/2 - 2 = -9/4`

Vertex is `(-1/2, -9/4)`

Axis of symmetry

Axis of symmetry is `x = -1/2`

x-intercepts

Equate the equation to zero and solve for 'x'

`x^2 + x - 2 = 0``x^2 + 2x - x - 2 = 0``(x + 2)(x - 1) = 0``x = -2 or x = 1`

x intercepts are (-2, 0) and (1, 0)

Points

• x = -3, `y = (-3)^2 + - 3 - 2 = 4`
• x = 2, `y = (2)^2 + 2 - 2 = 4`

Points are (-3, 4) and (2, 4)

Graph `y = x^2 + x - 2`