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`
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
(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.
Sketch the graph of the function `y = 2x^2− 8x + 6`
We first identify that `a = 2`, `b = -8` and `c = 6`.
Since `a = 2`, `a > 0` hence the function is a parabola with a minimum point and it opens upwards (U-shaped)
The x co-ordinate of the minimum point is:
The y value of the minimum point is
`y = 2(2)^2 - 8(2) + 6 = -2`
So the minimum point is `(2, -2)`
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.
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
Sketch the graph of the function `y = -x^2+ x + 6`
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
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
Get the Daily Math Tweet!
IntMath on Twitter
Sketch the graph `y = -x^2− 4x − 3`
`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