HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A QUADRATIC EQUATION

The graphical form of a quadratic function will be a parabola (u shpae). In which the maximum and minimum value will be there at vertex.

To find the maximum or minimum value from the quadratic equation, we have the following ways.

(i) Converting into the vertex form

(ii) Using formula

Find the maximum or minimum value of the following function.

Problem 1 :

y = –x² + 2x + 3

Solution :

Method 1 :

y = –x² + 2x + 3

Factoring negative sign.

y = –[x² - 2x - 3]

Write the coefficient of x as a multiple of 2.

y = –[x² - 2 ⋅ x ⋅ 1 + 1² - 1² - 3]

y = –[(x-1)² - 1 - 3]

y = –[(x-1)² - 4]

y = –(x-1)² + 4

It is exactly in the form of 

y = a(x - h)² + k

Here a = -1 < 0, then the parabola opens down.

Vertex is at (1, 4). So, minimum is at x = 1 and the minimum value = 4.

Method 2 :

y = –x² + 2x + 3

a = -1, b = 2 and c = 3

x = -b/2a

x = -2/2(-1)

x = 1

By applying the value of x in the given equation, we get

y = -(1)² + 2(1) + 3

y = -1 + 2 + 3

y = 4

So, the minimum point is at (1, 4).

Minimum value is y = 4

Problem 2 :

y = 2x² + 4x – 3

Solution :

y = 2x² + 4x – 3

Factoring 2, we get

y = 2(x + 1)² - 5

Here (h, k) is (-1, -5).

Vertex is at (-1, -5).

So,

minimum is at x = -1 and 

minimum value = -5.

Problem 3 :

y = –3x² + 4x 

Solution :

y = –3[x² + 4x/3]

Here (h, k) is (2/3, 4/3).

Vertex is at (2/3, 4/3).

So,

minimum is at x = 2/3 and 

minimum value = 4/3

Problem 4 :

x is the horizontal distance (in feet) and y is the vertical distance (in feet). Find and interpret the coordinates of the vertex.

a) The path of a basketball thrown at an angle of 45° can be modeled by

y = −0.02x² + x + 6

b) The path of a shot put released at an angle of 35° can be modeled by

y = −0.01x² + 0.7x + 6

Solution :

The path of a basketball thrown at an angle of 45°

y = −0.02x² + x + 6

y = −0.02[x² - x/0.02] + 6

= −0.02[(x - 1/0.04)² - (1/0.04)² ] + 6

= −0.02[(x - 1/0.04)² - (1/0.0016)] + 6

= −0.02(x - 1/0.04)² + (0.02/0.0016) + 6

= −0.02(x - 1/0.04)² + 12.5 + 6

= −0.02(x - 1/0.04)² + 18.5

Vertex is at (1/0.04, 18.5)

1/0.04 ==> 25

V (25, 18.5)

When basket ball reaches the maximum height 18.5 ft, the horizontal distance is 25 ft.

The path of a shot put released at an angle of 35°

y = −0.01x² + 0.7x + 6

y = −0.01[x² - 0.7x/0.01] + 6

= −0.01[x² - 70x] + 6

= −0.01[(x - 35)² + 35²] + 6

= −0.01[(x - 35)² + 1225] + 6

= −0.01(x - 35)² + 12.25 + 6

= −0.01(x - 35)² + 18.25

V (35, 18.5)

When basket ball reaches the maximum height 18.5 ft, the horizontal distance is 35 ft.

Problem 5 :

The engine torque y (in foot-pounds) of one model of car is given by

y = −3.75x² + 23.2x + 38.8

where x is the speed (in thousands of revolutions per minute) of the engine.

a. Find the engine speed that maximizes torque. What is the maximum torque?

b. Explain what happens to the engine torque as the speed of the engine increases.

Solution :

y = −3.75x² + 23.2x + 38.8

a) y = −3.75[x² - 6.18x] + 38.8

= −3.75[(x - 3.09)² + 3.09²] + 38.8

= −3.75[(x - 3.09)² + 9.54] + 38.8

= −3.75(x - 3.09)² - 35.775 + 38.8

= −3.75(x - 3.09)² + 3.13

The maximum speed of torque is 3.09 and maximum torque is 3.13.

b) When speed increases the engine torque will also increase.