Consider the function f(x) = x2 + 12x + 11. x-intercepts: 0 = x2 + 12x + 11 0 = (x + 1)(x+ 11) y-intercept: f(0) = (0)2 + 12(0) + 11 What are the intercepts of the function? The x-intercepts are . The y-intercept is

The coordinates of the x-intercepts are;

(-1, 0) and (-11, 0)

The coordinate of the y-intercept is;

(0, 11)

Step-by-step explanation:

The given function is f(x) = x² + 12·x + 11

The x-intercept is at f(x) = 0 which gives;

0 = x² + 12·x + 11 which has been factored as (x + 1)(x+11) such that we have;

(x + 1)(x+11) = 0

x  = -1, or x = -11

The y-intercept occurs at x = 0, we therefore have;

f(0) = (0)² + 12·(0) + 11 = 11

Therefore, we have f(0) = 11 = Y

Given that at the x-intercept, y = 0, then the coordinates of the x-intercepts are;

(-1, 0) and (-11, 0)

Given that at the y-intercept, x = 0, then the coordinate of the y-intercept is;

(0, 11).

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'.