Respuesta :

Answer:  The correct answer is:  

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The given "graph" in the bottom right, lowest corner

Step-by-step explanation:

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Note:  When there is only one (1) equation give for a graph;

          and/or:  only one (1)  "inequality given";  

         we  look for the symbol.

If the symbol is "not" an "equals" symbol (i.e. not an:  =  symbol) ;

 we check for the type of "inequality" symbol.

If there is a:  "less than" (<) ;  or a "greater than" (>) symbol;  the graph of the "inequality" will have "dashed lines" (since there will be a "boundary").

If there is an "inequality" that is a: "less than or equal to" (≤) ;

                                                 or a: "greater than or equal to" (≥) ;

       →   then there will be not be a dashed line when graphed;

          but rather—a "solid line" ;  since "less than or equal to" ;

          or "greater than or equal to" —is similar to:

          "up to AND including"; or: "lesser/fewer than AND including".).

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Note:  We are given the "inequality" :

            →   " y <  x²  + 4x  "  .

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Note that we have a "less than" symbol (< ) ; so the graph will have a:

   "solid line" [and not a "dotted line".].

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Note that all of the graphs among our 4 answer choices have "dotted lines".

Not that all values (all x and y coordinated) within the "shaded portion" of the corresponding graph are considered part of the graph.  

As such, given any point within the shaded part,  the x and y coordinates must match the inequality (i.e. the given inequality must be true when one puts in the "x-coordinate" and "y-coordinate" into the "given inequality" :

→   " y <  x²  + 4x "  .

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Likewise, we can take any point within the "white, unshaded" portion of any of the graph, and take the "x-coordinate" and "y-coordinate" of that point, and the inequality: →   " y <  x²  + 4x  "  ;  will not hold true when the "x-coordinate" and "y-coordinate" values of that point— are substituted into the "inequality".  

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{Note:  Answer is continued on images attached.}.

Wishing you the best!

Ver imagen JonHenderson55
Ver imagen JonHenderson55