What is the equation of the line that has a slope of -6/5 and goes through the point (5, -4)?

Step-by-step explanation:

Given

slope =6/5

point =(5, -4)

So

the equation of the line =

(y-y1)=m(x-x1)

Where X1=5,y1=-4,m=6/5

Put the values

(Y+4)=6/5(x-5)

5(Y+4)=6(x-5)

5y+20=6x-30

5y-6x=-30-20

-6x+5y=-50

6x-5y=50.

So the equation of the line is 6x-5y=50.

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