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Ksenya-84 [330]
3 years ago
13

How to find domain and range of a parabola?

Mathematics
1 answer:
german3 years ago
8 0
The domain is the value of x, the range is the value of y.
A parabola opens infinitely to the right and left, so x can be any number, the domain is all real numbers
Vertically, however, a parabola opens only one way, either upward or downward. when it opens upward from a a certain level, say, the lower point (the vertex) has a y coordinate of 2, we say the range is all real numbers larger than or equal to 2, or y≥2. When it opens downward, we say the range is all real numbers smaller or equal to 2, y≤2
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3x + y 2-3
sveticcg [70]

Answer:

The question is invalid. Please rewrite the question.

7 0
3 years ago
What describes the number and type of the roots of the equation 4x+7=0
Kaylis [27]

Answer:

D : One Real Root

Step-by-step explanation:

Isolate "4x" by subtracting 7 from both sides.

So we get

4x = -7

Then we divide each side by 4 to get -7/4

x = -7/4 so there is only one real root.

5 0
3 years ago
Graph the function.<br> f(x)=-\dfrac{2}{3}(x +1)(x -5)f(x)=− <br> 3<br> 2<br> ​<br> (x+1)(x−5)
Charra [1.4K]

Answer:

see below

Step-by-step explanation:

f(x) = -2/3 (x+1) (x-5)

First find the zeros

0 = -2/3 (x+1) (x-5)

Using the zero product property

x+1 =0   x-5=0

x= -1    x = 5

Then find the y intercept

y = -2/3 (0+1)(0-5)

  - 2/3 (1)(-5)

  +10/3

We know that this is a parabola that opens downward

6 0
3 years ago
20.<br>Find the slope of a line parallel to 3x - y = 1​
Olenka [21]

Answer:

m=3

m=Slope

Step-by-step explanation:

In order to find the slope of a line parallel to 3x-y=1

you must look at standard form form y=mx+b

the slope is m

so then m=3

6 0
3 years ago
Please help and thank you
ivann1987 [24]

First, let's try to isolate x to make the question easier to solve.

|2x-6|≤6

2x - 6 ≤ 6

2x - 6 ≥ -6

2x ≤ 12

x ≤ 6

2x ≥ 0

x ≥ 0

Knowing this, the graph should be equal or less than 6 and equal or greater than 0, making Option D the correct answer.

7 0
3 years ago
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