Question

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

Answer 1

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

Answer 2

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 the option B

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