Here we are going to see some example problems on graphing quadratic functions.

A quadratic function can be described by an equation of the form y = ax2 + bx + c, where a ≠  0.

 Open upward parabola Open downward parabola ## Opening of the parabola is on which side

(1)  (i) If the coefficient of x2 is positive, the parabola opens upward.

(ii) If the coefficient of x2 is negative, the parabola opens upward.

## Find the coordinates of the vertex.

This point, where the parabola changes direction, is called the "vertex".

We can find the x-coordinate of the vertex by using the formula, x = -b/2a

By applying the value of x in the given equation, we can get the y-coordinate value.

Vertex of the parabola (-b/2a, f(-b/2a)).

## Identify the vertex as a maximum or minimum.

• The parabola which opens downward has only the maximum value
• The parabola which opens downward has only the minimum value

## Draw the graph

By giving some random values of x, we can find the values of y.We can use the symmetry of the parabola to help us draw its graph. On a coordinate plane, graph the vertex and the axis of symmetry.

Let us look into some example problems to understand the above concept.

Example 1 :

Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Identify the vertex as a maximum or minimum. Then graph the function.

y = 2x2 - 4x - 5

Solution :

Axis of symmetry :

The given parabola is symmetric about y-axis. Since the coefficient of x2 is positive, the parabola opens upward direction.

Vertex of the parabola :

x  =  -b/2a

a = 2, b = -4 and c = -5

x  =  -(-4)/2(2)  ==>  4/4  ==> 1

By applying the value x = 1 in the given equation

y  =   2(1) - 4(1) - 5

=  2 - 4 - 5

=  -7

Hence the vertex of the parabola is (1, -7).

Identify the vertex as a maximum or minimum :

Since the parabola opens upward direction, it has only minimum value.

Draw the graph :

 x-2-1012 y111-5-7-5 Set of ordered pairs :(-2, 11) (-1, 1) (0, -5) (1, -7) and (2, -5) Example 2 :

Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of each function. Identify the vertex as a maximum or minimum. Then graph the function.

y = -3x2 - 6x + 4

Solution :

Axis of symmetry :

The given parabola is symmetric about y-axis. Since the coefficient of x2 is negative, the parabola opens downward direction.

Vertex of the parabola :

x  =  -b/2a

a = -3, b = -6 and c = 4

x  =  -(-6)/2(-3)  ==>  6/(-6)  ==> -1

By applying the value x = -1 in the given equation

y  =   -3(-1)2 - 6(-1) + 4

=  -3 + 6 + 4

=  -3 + 10

=  7

Hence the vertex of the parabola is (-1, 7).

Identify the vertex as a maximum or minimum :

Since the parabola opens downward direction, it has only maximum value.

Draw the graph :

 x-2-1012 y-474-5-20 Set of ordered pairs :(-2, -4) (-1, 7) (0, 4) (1, -5) and (2, -20)  After having gone through the stuff given above, we hope that the students would have understood "Graphing quadratic functions examples".

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