contestada

Which equation defines the graph of y=x^2 after it is shifted vertically 3 units down and horizontally 5 units left? A) y=(x-5)^2 - 3 B) y= (x+5)^2 - 3 c) y=(x-3)^2 - 5 D) y= (x+3)^2 - 5

Respuesta :

The answers is B because you set x+5=0 and x=-5 so that means that the graph needs to be shifted 5 units to the left because 5 is negative and to move 3 units down you just put a -3 at the end of the equation like this y=(x+5)-3
calculista

we have

[tex]y=x^{2}[/tex]

This is a vertical parabola open upward

the vertex is equal to the origin [tex](0,0)[/tex]

The rule of the translation is

[tex](x,y)------> (x-5,y-3)[/tex]

that means

the translations is [tex]5[/tex] units to the left and [tex]3[/tex] units down

therefore

the new vertex of the function will be

[tex](0,0)------> (0-5,0-3)[/tex]

[tex](0,0)------> (-5,-3)[/tex]

The new equation of the parabola in vertex form is equal to

[tex]y=(x+5)^{2}-3[/tex]

the answer is the option B

[tex]y=(x+5)^{2}-3[/tex]



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