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### 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)

### 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 = 0x^2 + 2x - x - 2 = 0(x + 2)(x - 1) = 0x = -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