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3 years ago

PLEASE HELP ASAP I WILL REWARD BRAINLIEST one unit up, one unit down, one unit left, one unit right,

Mathematics

2 answers:

labwork [276]3 years ago

8 0

Answer:

see below

Step-by-step explanation:

will make it simple.

1 unit up : y = |x+2| + 1

1 unit down : y = |x+2| - 1

1 unit to the left : y = |x + 3|

1 unit to the right : y = |x+1|

hope it helps. if its wrong, then please report it wo we can delete of correct it.

dusya [7]3 years ago

6 0

Answer:

y=x+2

Step-by-step explanation:

Since 1 up, 1 down, 1 left, and 1 right would all cancel out so you would still get y=x+2

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We have the following curve:

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To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.

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