Complete the equation of the line through (-6,5) and (-3,-3) y=

Please help I will give brainiest to the correct answer!

y=-8/3x-11 is the equation of the line and the value of y

Step-by-step explanation:

Use point slope form m = y-y1/x-x1, or y-y1=m(x-x1)

Find the slope from the two points (-6,5) and (-3,-3)

m = y2-y1/x2-x1 = -3-5/-3+6 = -8/3

use one point say (-6,5) that passes through the line and the slope to form the equation y-5=-8/3(x+6) rearrange y= -8/3(x+6)+5

y=-8/3x - 48/3 +5 = -8/3x-16+5 = -8/3x-11

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this