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

positive parabolanegative parabola                       

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 = {-1~-~2}/2  = -3/2

x = -3/2 is the axis of symmetry.

Applying this value of x in the quadratic equation, we get the value of y = f(-3/2) = -1/4

(-3/2,~-1/4) 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.

parabola (quadratic) function

Domain: {all real x}

Range: {y: y>=~-1/4

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 = {-1~+~5}/2  = 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.

negative (quadratic) function

Domain: {all real x}

Range: {y: y>=~9}