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1 year ago

What is the Y intercept? and where is the function positive?where is it negative?

Mathematics

1 answer:

kipiarov [429]1 year ago

8 0

y=2

$\lbrack0,1)\text{ }\cup\text{ (3,4}\rbrack$

$\begin{gathered} so,\text{ the function is negative in } \\ (1,3) \end{gathered}$

Explanation

Step 1

The y -intercept of a graph is the point where the graph crosses the y -axis.

so , let's check the graph

therefore, the y-intercept is 2

Step 2

The positive regions of a function are those intervals where the function is above the x-axis. It is where the y-values are positive (not zero)

hence, the function is potive in ( in x axis)

$\lbrack0,1)\text{ }\cup\text{ (3,4}\rbrack$

Step 3

negative region

The negative regions of a function are those intervals where the function is below the x-axis. It is where the y-values are negative (not zero).

$\begin{gathered} so,\text{ the function is negative in } \\ (1,3) \end{gathered}$

I hope this helps you

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but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

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