Consider the function
~f(x)=9-x^2~ f(x)=9−x 2 space, f, (, x, ), equals, 9, minus, x, squared, space for ~f(x) 0~ f(x)≥0 space, f, (, x, ), is greater than or equal to, 0, space only.
the shaded region is an isosceles triangle formed by joining the points ~(0,0), (0,0)space, (, 0, 0, ), ~big (x, f(x) big), (x, f(x))space, (, x, f, (, x, ), ), and ~big(-x, f(x) big) (−x, f(x))space, (, minus, x, f, (, x, ), ), where~0 x 3 0≤x≤3, 0, is less than or equal to, x, is less than or equal to, 3.
what is the area of the largest triangle that satisfies the stated conditions?

for the friend request.

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'.

step-by-step explanation:

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