If the point (3,5) is on f(x) find the corresponding point on the graph of y=(x+2)+1

1 answer:

8 0

The corresponding point to the point given (3,5) which is on the graph of f(x) as described in the task content is; (3,6).

<h3>What is the corresponding point to the given point (3,5) as given in the task?</h3>

It follows from the task content that the point given originally is; (3,5).

However, it is required that the point which is equivalent to the given point on the graph of y=(x+2)+1 be identified.

It therefore follows that the point required as in discuss is;

y = (3+2) +1

y = 6.

Hence, the corresponding point is; (3,6).

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pochemuha

Answer:

Down here ↓↓↓↓↓↓

Step-by-step explanation:

Anything below 1 is a reduce

Anything above 1 is an enlarge

Anything equal to 1 is a preserve

1. Reduce, 0.75 < 1

2. Reduce, 4/5 < 1

3. Enlarge, 4.2 > 1

4. Preserve, 8/8 = 1

5. Enlarge, 7/2 > 1

6. Preserve, 1 = 1

7. Reduce, 0.1 < 1

8. Enlarge, 1 2/3 > 1

9. Enlarge, 12 > 1

-Chetan K

3 0

2 years ago

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suter [353]

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4 0

2 years ago

Read 2 more answers

I need to find all real values of x from this inequality x^2>2x-2 and also this question x^2>2x+2. I need the answer fast

sergeinik [125]

$x^2\gt2x-2$

$x^2-2x+2\gt0$

You are essentially asking at this point where is the quadratic function greater than 0. To get the interval find the roots. Between the roots the function is guaranteed to be LESS THAN zero. Everywhere else the function is greater than 0.

$x^2-2x+2=0$

Use Vieta's rule to get the roots,

$(x-1)(x-1)=0$

Both roots are the same so no value is less than 0 which means all values except at the point of root are strictly greater, the point of root is at $x_{1,2}=1$ which means the set for which the quadratic function is greater than 0 is $\mathbb{R}-\{1\}$ "all real numbers except 1".