Maximum/Minimum Value of a Quadratic Expression

How is the graph of a quadratic equation? Before we discuss that, you must be familiar yourself with the standard form of quadratic equation :

f (x) = ax² + bx + c

The graph of a quadratic function is "U" shaped and is called a parabola. The exploration is carried by changing the values of all 3 coefficients a, h and k

                      

f(x) = a(x - h)² + k

The term (x - h)² is a square, hence either positive or equal to zero.

(x - h)² >= 0

If you multiply both sides of the above inequality by coefficient a, there are two possibilities to consider, a is positive or a is negative.

case 1: a is positive
a(x - h)² >= 0.

Add k to the left and right sides of the inequality

a(x - h)² + k >= k.

The left side represents f(x), hence f(x) >= k. This means that k is the minimum value of function f.

case 2: a is negative
a(x - h)² <= 0.

Add k to the left and right sides of the inequality

a(x - h)² + k <= k.

The left side represents f(x), hence f(x) <= k. This means that k is the maximum value of function f.

Note also that k = f(h), hence point (h,k) represents a minimum point when a is positive and a maximum point when a is negative. This point is called the vertex of the graph of f.

Example (Questions) : Find the vertex of the graph of each function and identify it as a minimum or maximum point.
a) f(x) = -(x + 2)² - 1
b) f(x) = -x² + 2
c) f(x) = 2(x - 3)²

a) f(x) = -(x + 2)² - 1 = -(x - (-2))² - 1
a = -1 , h = -2 and k = -1. The vertex is at (-2,-1) and it is a maximum point since a is negative.

b) f(x) = -x² + 2 = -(x - 0)² + 2
a = -1 , h = 0 and k = 2. The vertex is at (0,2) and it is a maximum point since a is negative.

c) f(x) = 2(x - 3)² = 2(x - 3))² + 0
a = 2 , h = 3 and k = 0. The vertex is at (3,0) and it is a minimum point since a is positive.

Equations of Straight Lines Part 1

(i) Slope-intercept form 

Equation of straight line with slope m and y - intercept c is given by: 

                                               y = mx + c

Example

Find the equation of the line with slope is 2 and c=5.

Solution

                                                             y = mx + c

  so                                                        y = 2x + 5

(ii) Point-slope form 

Let m be a slope of line and A(x y 1 1 , ) be a point lies on a line then equation of
line is given by:

 

                                               y - y₁ = m(x - x₁)

Example

Find the equation of the line with gradient 2 passing through (1, 4).

Solution

y - 4 = 2(x - 1)

y - 4 = 2x - 2

y = 2x + 2