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

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

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

Answer 2

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