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9966 [12]
4 years ago
11

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

or anyone who helps!
Y=(x-5)^2-3
Y=(x+5)^2-3
Y=(x-3)^2-5
Y=(x+3)^2-5
Mathematics
2 answers:
bazaltina [42]4 years ago
8 0

Answer:

B. y=(x+5)^{2} -3

Step-by-step explanation:

we have y=x^{2}. This is a vertical parabola open upward the vertex is equal to the origin (0,0). The rule of the translation is (x,y)⇒(x-5,y-3) that means the translations is 5 units to the left and 3 units down. Therefore, the new vertex of the function will be (0,0)⇒(0-5,0-3); (0,0)⇒(-5,-3). The new equation of the parabola in vertex form is equal to y=(x+5)^{2} -3. The answer is B. y=(x+5)^{2} -3

Kay [80]4 years ago
4 0
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
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