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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

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

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

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