Theres a 4 at the bottom but i couldn't get it in

See answer below.

Step-by-step explanation:

1. Start at point (0,0), This point is where the x-y-axis meet.  Do not put a dot

   there because this is your starting point.

2.  Move to the left over to point -1 3/4 . This place is between -1 and -2, and

     closer to -2 because the point is -1 .75.  Remember this point is not on the

     graph.  

3.  Now move up from that point to 4.5 and put a dot at that place.  The point is in the Q II ,  to the left top.  

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

1/2*qr*rs

step-by-step explanation: