which equatio defines the graph of y=x^2 after it is shifted veritcally 3 units down and horizontally 5 units left
The equation that defines the graph of y = x^2 after it is shifted vertically 3 units down and horizontally 5 units left is:
y = (x + 5)^2 - 3
To shift the graph of y = x^2 vertically 3 units down, we subtract 3 from the equation. The shifted equation becomes y = x^2 - 3.
To shift the graph of y = x^2 horizontally 5 units left, we replace x with (x + 5) in the equation. The shifted equation becomes y = (x + 5)^2 - 3.
Therefore, the equation that defines the graph of y = x^2 after it is shifted vertically 3 units down and horizontally 5 units left is y = (x + 5)^2 - 3.
To shift the graph of the equation y = x^2 vertically 3 units down and horizontally 5 units left, you can use the general form of the quadratic equation with the horizontal and vertical shifts:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola, and the values a, h, and k determine the shifts in the horizontal and vertical directions.
In this scenario, the vertical shift is 3 units down, which means k = -3. The horizontal shift is 5 units left, so h = -5. We don't have any information about the value of a (the coefficient in front of the squared term), so we'll assume it remains the same as in the original equation, which is usually 1.
Now, let's substitute the values into the equation:
y = 1(x - (-5))^2 + (-3)
Simplifying further:
y = (x + 5)^2 - 3
Therefore, the equation that defines the graph of y = x^2 after it is shifted vertically 3 units down and horizontally 5 units left is y = (x + 5)^2 - 3.