Question

Hi! Can someone help me with these? Thanks!


1.) Determine the slope and the y-intercept of the equation of the line passing through the points (-5, 3) and (1, -3). Write the equation of the line in slope intercept form.



2.) Determine the x- and y- intercepts and use them to graph the following equation. y=x^2-2x-3

Answer 1

(1)
The slope is -6/6 = -1
So, in point-slope, we have
y-3 = -1(x+5)
Now massage that into slope-intercept form.

Clearly, when x=0, y = -3
Since y = (x-3)(x+1), y=0 when x is -1 or 3.
So, draw a parabola with those roots, which opens upward. Recall that the vertex occurs midway between the roots, where x = 1.

Answer 2

slope = (-3 - 3)/ (1+5) = -6/6 = -1
y = -x + b
3 = -1(-5) + b
b = -2

y = -x - 2
===============================
x axis(0, -3)

y axis where y = 0
x^2 - 2 x - 3 = 0
(x-3)(x+1) = 0
x = 3 or x = 1
so
axis of symmetry and vertex at x = (1+3)/2 = 2
then y vertex = 4-4 - 3 = -3
so vertex at (2 , -3) opens up
through (1,0) and (3,0)

Answer 3

dumb error