Dear Ms. Quadratic I followed all of the directions given in your
brochure on how to order a parabola. I tried to
order a parabola that is passed through the points
(1,0) and (-6,0)

We have the parabola as;

x^2 + 5x - 6

Step-by-step explanation:

Here, we want to get a parabola that passes through the points (1,0) and (-6,0)

The general form of a quadratic equation is;

y = ax^2 + bx + c

We have two x-intercepts at x = 1 and x = -6

A quadratic equation is expected to have two roots only and that means we have these two values as the roots

Hence, we have ;

(x-1) and (x + 6)

The parabola is thus;

(x-1)(x + 6)

= x(x + 6) -1(x + 6)

= x^2 + 6x - x -6

= x^2 + 5x - 6

step-by-step explanation:

if u want da anser kill your teeser ifu hav wun

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'.