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3 years ago

How do you find the minimum value of a quadratic function?

Mathematics

1 answer:

aniked [119]3 years ago

6 0

The minimum value of a function is the place where the graph has a vertex at its lowest point.

There are two methods for determining the minimum value of a quadratic equation. Each of them can be useful in determining the minimum.

(1) By plotting graph

We can find the minimum value visually by graphing the equation and finding the minimum point on the graph. The y-value of the vertex of the graph will be the minimum.

(2) By solving equation

The second way to find the minimum value comes when we have the equation y = ax² + bx + c.

If our equation is in the form y = ax^2 + bx + c, you can find the minimum by using the equation min = c - b²/4a.

The first step is to determine whether your equation gives a maximum or minimum. This can be done by looking at the x² term.

If this term is positive, the vertex point will be a minimum; if it is negative, the vertex will be a maximum.

After determining that we actually will have a minimum point, use the equation to find it.

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Express decimal as a fraction 8.25

aksik [14]

Answer:

33/4

Step-by-step explanation:

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2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x+2}$ if it is defined for $x = 0$ .

Answer:

$y = \sqrt {x-2}$

Option b

6 0

3 years ago

Can you please help me!!

Molodets [167]

A is more because 1/4 is equal to 2/8 and 2/8 is less than 7/8 so you would have more than a minute
B is three minutes 2/8 goes into 7/8 three times

8 0

3 years ago

Write perimeter and area of circle

alexira [117]

Answer:

The perimeter of a circle is π × d. Here's is the diameter of the circle. Hence, the perimeter of a circle is half of that of the circle that is ½ π × d.

The area of a circle is pi times the radius squared (A = π r²). Learn how to use this formula to find the area of a circle when given the diameter.

Step-by-step explanation:

8 0

3 years ago

Is 40 a multiple or a factor

gladu [14]

Hey there!
It can be both. It's a multiple of 8 because 8 times 5 is 40, but it's also a factor of 120 because 80 times 3 is 120.
Hope this helps!

5 0

2 years ago