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

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

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