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alexandr402 [8]
3 years ago
9

Which line(s) intersect the parabola y = x^2 − 3x + 4 at two points? Select all that apply. A. y = –3x + 2 B. y = –3x + 3 C. y =

–3x + 5 D. y = –3x + 6
Mathematics
1 answer:
TEA [102]3 years ago
4 0

Answer:

C and D

Step-by-step explanation:

Equating the line A and the parabola, we get

-3x + 2 = x² - 3x + 4

0 = x² - 3x + 4 +3x - 2

0 = x² + 2

-2 = x²

which has no real solutions. Then, the line A and the parabola don't intersect each other.

Equating the line B and the parabola, we get

-3x + 3 = x² - 3x + 4

0 = x² - 3x + 4 + 3x - 3

0 = x² + 1

-1 = x²

which has no real solutions. Then, the line B and the parabola don't intersect each other.

Equating the line C and the parabola, we get

-3x + 5 = x² - 3x + 4

0 = x² - 3x + 4  + 3x - 5

0 = x² - 1

1 = x²

√1 = x

which has 2 solutions, x = 1 and x = -1. Then, the line C and the parabola intersect each other.

Equating the line D and the parabola, we get

-3x + 6 = x² - 3x + 4

0 = x² - 3x + 4  + 3x - 6

0 = x² - 2

2 = x²

√2 = x

which has 2 solutions, x ≈ 1.41 and x ≈ -1.41. Then, the line D and the parabola intersect each other.

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What is the range for the following function? y=1/(x+2)+3
LUCKY_DIMON [66]

Answer:

\large \boxed{\text{C) }\{y: y \in \mathbb{R}, y \ne 3\} }

Step-by-step explanation:

The range is the spread of the y-values (minimum to maximum distance travelled).

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There is a vertical asymptote at x = -2, so y does not exist when x = -2.  However,

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Since the limit from either side is the same,

\displaystyle \lim_{x \rightarrow -{2}}f(x) = 3

The graph below shows the asymptotes of your function.

Thus. y can take any value  except 3.

In set builder notation, the range is  

\large \boxed{\mathbf{\{y: y \in \mathbb{R}, y \ne 3\} }}

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