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Graphing quadratic in standard form

The method of graphing a quadratic when it is in the standard form is described in four steps:
  1. Get the coordinates of the vertex of the parabola
  2. Get the axis of symmetry
  3. Get the x-intercepts or some points coordinates
  4. Plot the vertex and x-intercepts/points and join them with a free hand curve
    We will discuss the above two steps with the help of the example:
    `y = 2x^2  + 3x + 1`

    Get the vertex of the parabola

    Let the vertex of the parabola be `(h, k)`, where 'h' is the x-coordinate, and 'k' is the y-coordinate. We will get the value of the x-coordinate from the following formula:
    `h = -b/(2a)`
    On comparing `2x^2 + 3x + 1` with `ax^2 + bx + c`, 
    • `a = 2`
    • `b = 3`
    • `c = 1`
    Therefore in the quadratic `2x^2 + 3x + 1`, the x-coordinate of the vertex is
    `h = -3/4`
    To get the value of the y-coordinate of the vertex (k), substitute the x-coordinate (obtained above) `-3/4` into the quadratic function. The value computed by this will be the y-coordinate of the vertex. In our example, we get
    `k = 2(-3/4)^2 + 3(-3/4) + 1`
    `k = -1/8`
    Therefore the vertex is `(-3/4, -1/8)`.

    Get the axis of symmetry

    Although you do not necessarily need to graph the axis of symmetry, it is useful to know the axis of symmetry of a parabola when graphing it.
    The axis of symmetry is the x-coordinate of the vertex. Since the vertex is `(-3/4, -1/8)`, hence the axis of symmetry is `x = h` that is `x = -3/4`

    Get the x-intercepts or some points coordinates

    Apply the quadratic formula to get the x-intercepts
    `x = (-b ± √(b^2 - 4ac))/(2a)`
    `x = (-3± √ ((-3)^2 - 4(2)(1)))/(2*2)`
    `x = (-3± 1)/4`
    `x = -1/2, x = -1`
    x-intercepts are written in the form `(x, 0)`. So the x-intercepts are `(-1/2, 0)` and `(-1, 0)`.

    Note that some quadratic functions do not have x-intercepts and some quadratic functions have the x-intercept and vertex at the same point. In this case, you need to get the coordinates of two points on either side of the axis of symmetry. For example, if the quadratic function above did not have x-intercepts or its x-intercept coincided with the vertex, then you would proceed as follows:

    Take one or more x-values on the left side (lesser than) the axis of symmetry and substitute them in the equation to get the y-coordinates of those points:

    • `x = -2, y = 2(-2)^2 + 3(-2) + 1 = 3` - point is `(-2, 3)`
    Take one or more x-values on the right side (greater than) the axis of symmetry and substitute them in the equation to get the y-coordinates of those points:
    • `x = 1, y = 2(1)^2 + 3(1) + 1 = 6` - point is `(1, 6)`

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

    The following graph of `y = 2x^2 + 3x + 1` is obtained. The vertex, x-intercepts and points obtained in steps 1 to 3 above are marked on the graph by dots:

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