Question

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

Answer 1

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

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`

Answer 2

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