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Which equation defines the graph of y=x^2 after it is shifted vertically 3 units down and horizontally 5 units left? Thank you for anyone who helps!

Y=(x-5)^2-3
Y=(x+5)^2-3
Y=(x-3)^2-5
Y=(x+3)^2-5

Respuesta :

HomertheGenius
The parent function is : y = x².
The new equation:  y = a( x - k )² + h,
where:  a = 1,  k = - 5 ( shifted horizontally 5 units left ),
h = - 3 ( shifted vertically 3 units down ).
Answer:
A ) y = ( x + 5 )² - 3
QueasyUnicorn14

Answer:

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

Step-by-step explanation:

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

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