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kupik [55]
3 years ago
12

Which of the following functions is graphed below.

Mathematics
1 answer:
Valentin [98]3 years ago
7 0

Answer: Option A

y=\left \{ {{x^2 +2;\ \ x

Step-by-step explanation:

In the graph we have a piecewise function composed of a parabola and a line.

The parabola has the vertex in the point (0, 2) and cuts the y-axis in y = 2.

The equation of this parabola is y = x ^ 2 +2

Then we have an equation liney = -x + 2


Note that the interval in which the parabola is defined is from -∞ to x = 1. Note that the parabola does not include the point x = 1 because it is marked with an empty circle " о ."

(this is x< 1)

Then the equation of the line goes from x = 1 to ∞ . In this case, the line includes x = 1 because the point at the end of the line is represented by a full circle .

(this is x\geq 1)

Then the function is:

y=\left \{ {{x^2 +2;\ \ x

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What are the intercepts of x^2+y-9=0
Dmitry_Shevchenko [17]

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Zeros

x = -3   point (-3 ,0)

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Step-by-step explanation:

Using a graphing calculator, we can easily check the intercepts of the equation

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Hi help with these please and thank yu? :)
ryzh [129]

Answer:

the first one is A

b/c when you calculate it comes 0 = 6

And the second one is D

b/c when you calculate it comes 0 = 2

if it's helpful ❤❤❤

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7 0
3 years ago
I don't know how to solve it.
Nataly [62]

Step-by-step explanation:

Take the first derivative

\frac{d}{dx} ( {x}^{3}  - 3x)

3 {x}^{2}  - 3

Set the derivative equal to 0.

3 {x}^{2}  - 3 = 0

3 {x}^{2}  = 3

{x}^{2}  = 1

x = 1

or

x =  - 1

For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).

For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.

Since we are going to Positve to negative slope, we have a local max at x=-1

Plug in -1 for x into the original function

( - 1) {}^{3}  - 3(  - 1) = 2

So the local max is 2 and occurs at x=-1,

For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.

Since we are going to decreasing to increasing, we have minimum at x=1,

Plug in 1 for x into original function

{1}^{3}   - 3(1)

1 - 3 =  - 2

So the local min occurs at -2, at x=1

8 0
2 years ago
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