Starting with the parent function f(x) = sq rt x. (where sq rt is "the square root of" . I always remember that vertical shifts are "tagged on the end" and horizontal shifts are part of what is happening to x, because the x values are along the horizontal axis (x-axis). So a horizontal shift of 6 to the left, is going to be x + 6. It is the opposite of what you think here, adding moves the graph left and subtracting moves the graph right. Now, we have sq rt (x+6) => the sq rt function shifted left. We still need to move the graph down 5 units. Since this is a change in the location along the y-axis, we just subtract (because it is moved down) from the term sqrt(x+6). So we now have f(x) = sqrt(x+6) -5.
The function is y = sqrtx Or y = x^(1/2).
If the transformation is a shift of (h, k), then the equation becomes:
X = x+h and Y = x+k and the new equation is
Y-k =(X-h)^(1/2) Or
y-k = (x-h)^(1/2). Therefore,
(y-k)^2 = (x-h) is a parabola.
The vertex is at ( h, k) or (-6, -5)
Also y^2-2yk+ k^2-h -x = 0 is a quadratic in x with a discriminant, 4k^2-4(k^2+h-x) = 4(x-h) .
Since the discriminant is -h or +5 as h =5. Therefore parabola intercepts Y axis at two points k+sqrt(0+h) and k+sqrt(0+h) or at -5+sqrt6 or -5-sqrt6
y = k Or y = -5 or y+5=0 is the axis of symmetry.
Firther, (y-k)^2 = 4(1/4)(x-h) . 2(1/4) or 1/2 is the latus rectum of the parabola which is independent of (h,k).
When y = 0 in (y-k)^2 = x-h, x=k^2+h Or x = (5)^2-6 = 19 is the intercept of x axis.
The focus of the parabola is 1/4 units right of h. Or the coordinates of the focus is (h+1/4, k) Or
(-6+1/4, -5) = (-5.75, -5).
Thus the parabola ocupies all 4 quadrarnts, with its vertex at (-6,-5), focus (-5.75, -5) , axis of symmetry y +5 = 0, latus rectum 1/2, only one interceptsof x axis at 19 units and two y intercepts at -5+sqrt6 and -5+sqrt6. The parabola opens to the right and is compressed along x axis.