For items 5 - 8, identify the quadratic functions using the given a graph. A. y =(x - 7)2 – 6 B. y =(x + 7)² +6 C. y =(x + 7)2 - 6 D. y = (x – 7)² +6 5. fo 12 14
Minimization
In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in terms of global or local range. The definition of minimization in the thesaurus is the process of reducing something to a small amount, value, or position. Minimization (noun) is an instance of belittling or disparagement.
Maxima and Minima
The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.
Derivatives
A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which one would be easier? Walking on the road would be much easier than climbing a mountain.
Concavity
In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the first derivative test and second derivative test to understand the concave behavior of the function.
![For items 5 - 8, identify the quadratic functions using the given a graph.
A. y = (x – 7)2 – 6
B. y =(x + 7)2 + 6
C. y =? (x + 7)2 – 6
5.
D. y =(x – 7)2 +6
1
A. y = (x – 2)² + 2
B. y = -(x – 2)² + 2
C. y = (x – 2)2 + 2
D. y = -(x – 2)² – 2
6.
2
-2
10
A. y = (x – 2)
B. y = -(x – 2)?
C. y =(x + 2)2
D. y = -(x + 2)?
7.
A. y = -4(x + 2)² + 1
B. y = -4(x – 2)² – 1
C. y = 4(x – 2)² + 1
D. y = 4(x – 2)² – 1
8.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe77e1d92-78bf-43e4-9431-79cdba19ba8e%2F2226661b-3198-4eac-a846-7dfba867b012%2Fxqr975a_processed.jpeg&w=3840&q=75)
![• Follow This One!
Study the graph of the quadratic function below. Determine the equation
given in the graph by following the steps below.
1. Identify the vertex (h, k)
2. Identify the coordinates of
any point on the parabola.
3. Substitute the vertex (h, k)
and coordinates of any
point (x, y)into the vertex
form y = a(x – h)² + k.
4. Get the value of a.
5. Write the equation of the
quadratic function.
Example
• Try to Understand!
When the vertex and any point on the parabola are clearly seen, the
equation of the quadratic function can be determined by using the form of a
quadratic function y = a(x – h)2 + k.
Do the "Follow This One!" activity above. The vertex of the graph of
the quadratic function is (4, -2). The graph passes through the point (2,0). By
11
replacing x and y with 2 and 0, respectively, and h and k with 4 and -2
respectively, we have
y = a(x – h)² + k
(0) = a(2 – 4)2 + (-2) the x-intercept
0 = 4a – 2
ta
Substitute the vertex and
Substitute the vertex, h=4, k=.
2 and the value of a= to y =
Simplify
Divide both sides
a(x – h)² + k form the
required quadratic function
y = a(x – h)² + k
by 4
1
y =; (x – 4)2 – 2
Thus, the quadratic equation is y = (x – 4)² – 2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe77e1d92-78bf-43e4-9431-79cdba19ba8e%2F2226661b-3198-4eac-a846-7dfba867b012%2F6xo3tll_processed.jpeg&w=3840&q=75)
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