Home > Introduction to Pre-Calculus > Introduction to Graphing Functions > Quadratic Functions

The most common form of graphing quadratic functions results in a parabola, one of the most common types of quadratic functions.

f(x) = ax2 + bx + c is a quadratic equation, and it represents the equation for a parabola.

If a>0 (a is positive), the parabola is concave upwards, and if a<0 (a is negative), the parabola is concave downwards.

Parabolas have widespread applications in day to day lives, e.g. car head lights for wider diffusion of light, and satellite dishes to receive signals.

Example 1: Sketch the graph of y = f(x) = x2 + 3x + 2, and state the domain and range.

Step 1: First we need to find the x and y intercepts. For x-intercepts, y = 0.

x2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = -1 or -2

These points are (-1, 0) and (-2, 0) are called zeros rather than x-intercepts.

Next, to find y-intercepts, x = 0.

y = f(x) = 02 + 3×0 + 2  =  2

(0, 2) is the y-intercept where the graph meets y-axis.

Step 2: Since the quadratic equation y = f(x) = x2 + 3x + 2 has a>0, the parabola will be concave up.

Step 3: Now draw the graph/curve using the three points in a curve which is concave upwards. To sketch the graph better, we follow one more step, by finding the axis of symmetry of the curve.

The parabola has a minimum value, since a>0. The parabola is symmetrical, so the axis of symmetry lies halfway between the x-intercepts (zeros).

x =  =

x =  is the axis of symmetry.

Applying this value of x in the quadratic equation, we get the value of y =  =

is the minimum point of the parabola, also called the vertex of the parabola.

Step 4: Now draw the curve through the x-intercepts, y-intercepts and the vertex.

Domain: {all real x}

Range: {y:

Example 2: Sketch the graph of y = f(x) = -x2 + 4x + 5, and state the domain and range.

For x-intercepts, y = 0

y = -x2 + 4x + 5 = 0

x2 – 4x – 5 = 0

(x – 5)(x + 1) = 0

x = 5, or x = -1.

x-intercept points are (-1, 0) and (5, 0)

For y-intercept, x = 0. So

f(0) = -(0)2 + 4×0 + 5  = 5.

y-intercept is (0, 5)

Since a<0, the parabola will concave downwards, so it will have a maximum value.

Axis of symmetry =  = 2.

To find the vertex – x = 2, so y = f(2) = -(2)2 + 4×2 + 5  = -4 + 8 + 5  = 9

Maximum value = 9

Vertex = (2, 9)

Now sketch the parabola with x-intercepts (-1, 0) and (5, 0), y-intercepts (0, 5) and vertex (2, 9) with concave downwards.

Domain: {all real x}

Range: {y: }