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Arturiano [62]
3 years ago
5

Forms of Quadratic equations Please help me out

Mathematics
1 answer:
julia-pushkina [17]3 years ago
4 0

<em>Standard form:</em>

1. y = x² - 4x - 5

2. y = x² - 2x - 8

3. y = x² + 4x - 21

4. y = x² + 12x + 20

5. y = x² + 8x + 12

6. y = x² + 4x + 3

7. y = x² - 8x + 7

8. y = x² - 6x + 8

9. y = x² + 10x - 11

<em>Vertex</em><em> </em><em>form</em><em>:</em>

1. (x - 2)² - 9

2. ----

3. (x + 2)² - 25

4. ----

5. (x + 4)² - 4

6. (x + 2)² - 1

7. ----

8. (x - 3)² - 1

9. (x + 5)² - 36

<em>Factored form</em>

1. (x + 1) (x - 5)

2. (x + 2) (x - 4)

3. (x - 3) (x + 7)

4. (x + 10) (x + 2)

5. (x + 6) (x + 2)

6. (x + 3) (x + 1)

7. (x - 7) (x - 1)

8. (x - 2) (x - 4)

9. (x + 11) (x - 1)

<em>Questions</em>

1. Factored form

2. Factored form

3. Vertex form

4. Standard form

5. Standard form

6. Vertex form

NB: Finally finished ^-^ I hope all the answers to the <em>questions</em><em> </em>are correct.

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Answer:

C arithmetic

Step-by-step explanation:

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To get from the first term to the second term, we add 2

To get from the second term to the third term, we add 2

To get from the third term to the fourth term, we add 2

The common difference is 2

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Step-by-step explanation:

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Step-by-step explanation:

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Yakvenalex [24]
For f to be continuous at x=1, you need to have the limit from either side as x\to1 to be the same.

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2
\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b

If a=2 and b=3, then the limit from the right would be 2+3=5\neq2, so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation a+b=2.

Part (c) is done similarly to part (1). This time, you need to limits from either side as x\to2 to match. You have

\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b
\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0

So, a and b have to satisfy the relation 4a+2b=0, or 2a+b=0.

Part (4) is done by solving the system of equations above for a and b. I'll leave that to you, as well as part (5) since that's just drawing your findings.
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