20 points thanks :) im kinda stuck lol

1. (a) 0 = x² + 4x - 12 (Switch Sides)

x² + 4x - 12 = 0 (Factor Expression)

(x - 2)(x + 6) = 0 (Apply the Zero Factor Principle)

x = 2, and x = - 6

x - intercept : (2, 0) and ( - 6, 0)

(b) y = (0)² + 4(0) - 12 (Simplify)

y = 0 + 4 * 0 - 12 = 0 + 0 - 12 = - 12

y - intercept : (0, - 12)

(c) Our vertex belongs to an upward facing parabola, and hence it will be a minimum, of ( - 2, - 16).

2. Seth's solution is not accurate. He did the calculations accurately, and ended up with x = 3 and x = - 2. However he forgot to substitute those values back into the equations y = x² + 3x - 5 and y = 4x + 1, solving for the y values. Instead of being ( - 2, 0) and (3, 0) it should have been (3, 13), ( - 2, - 7)...

y = (3)² + 3(3) - 5 = 9 + 9 - 5 = 18 - 5 = 13,

y = 4( - 2) + 1 = - 8 + 1 = - 7

Hence we get the solutions (3, 13) and ( - 2, - 7).

answer: okw

step-by-step explanation:

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'.