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4. The Graph of the Quadratic Function

In general, the graph of a quadratic equation

`y = ax^2+ bx + c`

is a parabola.

[You can also see a more detailed description of parabolas in the Plane Analytic Geometry section.]

Shape of the parabola

If `a > 0`, then the parabola has a minimum point and it opens upwards (U-shaped) eg.

`y = x^2+ 2x − 3`

If `a < 0`, then the parabola has a maximum point and it opens downwards (n-shaped) eg.

`y = -2x^2+ 5x + 3`

Sketching Parabolas

In order to sketch the graph of the quadratic equation, we follow these steps :

(a) Check if `a > 0` or `a < 0` to decide if it is U-shaped or n-shaped.

(b) The Vertex: The x-coordinate of the minimum point (or maximum point) is given by

`x=-b/(2a)`

(which can be shown using completing the square method, which we met earlier).

We substitute this x-value into our quadratic function (the y expression). Then we will have the (x, y) coordinates of the minimum (or maximum) point. This is called the vertex of the parabola.

(c) The coordinates of the y-intercept (substitute `x = 0`). This is always easy to find!

(d) The coordinates of the x-intercepts (substitute `y = 0` and solve the quadratic equation), as long as they are easy to find.

Example 1

Sketch the graph of the function `y = 2x^2− 8x + 6`

Answer

We first identify that `a = 2`, `b = -8` and `c = 6`.

Step (a)

Since `a = 2`, `a > 0` hence the function is a parabola with a minimum point and it opens upwards (U-shaped)

Step (b)

The x co-ordinate of the minimum point is:

`x=b/(2a)=(-(-8))/(2(2))=8/4=2`

The y value of the minimum point is

`y = 2(2)^2 - 8(2) + 6 = -2`

So the minimum point is `(2, -2)`

Step (c)

The y-intercept is found by substituting `x = 0` into the y expression.

`y = 2(0)^2 - 8(0) + 6 = 6`

So `(0, 6)` is the y-intercept.

Step (d)

The x-intercepts are found by setting `y = 0` and solving:

`2x^2 - 8x + 6 = 0`

`2(x^2 - 4x + 3) = 0`

`2(x - 1)(x - 3) = 0`

So `x = 1`, or `x = 3`.

Using the above information, the sketch of the curve will be :

`y = 2x^2 -8x + 6`, a U-shaped parabola, showing intersections with axes

Example 2

Sketch the graph of the function `y = -x^2+ x + 6`

Answer

We first identify that `a = -1`, `b = 1` and `c = 6`

Step (a) Since `a = -1`, `a < 0` hence the function is a parabola with a maximum point and it opens downwards (n-shaped)

Step (b) The x co-ordinate of the maximum point is

`x=-b/(2a)=(-(1))/(2(-1))=(-1)/-2=1/2`

So

`y=-(1/2)^2+1/2+6=6 1/4`

Step (c) The y-intercept is the point `(0, c) = (0, 6)`

Sometimes, we need some more points to get a better sketch of the parabola.

Two points we can also find are the x-intercepts ie. points where the function cuts the x axis (where `y = 0`).

To find the x intercepts, we let `y = 0` and we get `-x^2 + x + 6 = 0`

or `x^2 - x - 6 = 0`

Factoring `(x - 3) (x + 2) = 0`

Solving `x - 3 = 0` or `x + 2 = 0`

So `x = 3` or `x = -2`.

Hence, the x intercepts are `(3, 0)` and `(-2, 0)`.

Here is our sketch:

`y = -x^2 + x + 6`, showing maximum point and intersections with axes

Exercise

Sketch the graph `y = -x^2− 4x − 3`

Answer

`y = -x^2 - 4x - 3`

`a < 0` so it is n-shaped.

Max value is when `x=-(-4)/-2=-2`

So the maximum point is `(-2, 1)`

`-x^2 - 4x - 3 = 0` when

`x^2 + 4x + 3 = 0`

`(x + 1)(x + 3) = 0 `

So x-axis intercepts are `x = -1` and `x = -3`.

`y = -x^2 -4x -3`, showing maximum point and intersections with axes

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