Given that h(x) = - (x - 1)^2 - 1, write an expression for f(x) in terms of x.
f(x) =

f(x) = -(x-1)^2+5

Step-by-step explanation:

If you look at the vertex of both graphs then f(x) is 6 units higher than h(x)

so if you go up 6 units then you just add 6 to the function

f(x) = h(x) + 6

f(x) = -(x-1)^2 - 1 + 6

f(x) = -(x-1)^2+5

Provided by guathmaths

f(x) = -(x-1)^2+5

Explanation:

f(x) = h(x) + 6

f(x) = -(x-1)^2 - 1 + 6

f(x) = -(x-1)^2+5

Must click thanks and mark brainliest

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when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.