**Maxima, minima and zeros of quadratic functions Multiplication of Decimals**

Let f(x) = ax2+bx+c represent a quadratic function.

If f(x) = 0

= ax2+bx+c = 0

Then, the **roots** of the quadratic equation are also called the **zeros** of the quadratic function.

If f(x)=ax2+bx+c gradually increases until it reaches a value p, which is algebraically greater than its neighboring values on either side, then p is said to be the **maximum value** or **maxima** of f(x).

If f(x)= ax2+bx+c gradually increases until it reaches a value q, which is algebraically less than its neighboring values on either side, q is said to be the minimum value or minima of f(x).

These concepts can be visualized by drawing graphs.

**Example : 1**

Draw the graph for x2+x-6 and find its zeros, maxima and minima, if they exist.

y = x2+x-6

x |
-4 |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |

x2 |
16 |
9 |
4 |
1 |
0 |
1 |
4 |
9 |

x |
-4 |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |

-6 |
-6 |
-6 |
-6 |
-6 |
-6 |
-6 |
-6 |
-6 |

y |
-6 |
0 |
-4 |
-6 |
-6 |
-4 |
0 |
6 |

Plot the points (-4,6), (-3,0), (-2,-4), (-1,-6), (0,-6),(1,-4),(2,0),(3,6) and join them to form a smooth curve.

From the graph, the curve cuts the x-axis at (-3,0) and (2, 0).

So the zeros of x2+x-6 = 0 are x = -3,2.

The maxima does not exist, as the curve continues indefinitely on both sides.

The minima is obtained by drawing a line from point L to M on the x-axis.

LM=-6.25, which is the minimum value or minima and it occurs at x=-0.5.

Draw the graph for the function f(x)= 4-5x-x2. From the graph obtain the zeros, maxima and minima, if they exist.

Let y = 4-5x-x2

x |
-6 |
-5 |
-4 |
-3 |
-2 |
-1 |
0 |
1 |
2 |

-x2 |
-36 |
-25 |
-16 |
-9 |
-4 |
-1 |
0 |
-1 |
-4 |

-5x |
30 |
25 |
20 |
15 |
10 |
5 |
0 |
-5 |
-10 |

4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |
4 |

y |
-2 |
4 |
8 |
10 |
10 |
8 |
4 |
-2 |
-10 |

Plot the points (-6,-2), (-5,4),(-4,8),(-3,10),(-2,10),(-1,8), (0,4),(1,-2),(2,-10) and join them to form a smooth curve.

From the graph, we see that the curve cuts the x-axis at the points (-5.7,0) and (0.7,0). So the zeros of 4-5x-x2 = 0 are x = -5.7, 0.7

Let L be the point where the curve is maximum. Drop a perpendicular from L to the x-axis, meeting the x-axis at M.

LM = 10.25. This is the maximum value and it occurs at

x = -2.5.

There is no minimum value as the curve continues on indefinitely on both sides.

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