The area of triangle formed by points of intersection of parabola y=a(x−1)(x−4) with the coordinate axes is 3. find a if it is known that parabola opens downward.

  a = -1/2

Step-by-step explanation:

In order for the triangle with a base of 4-1 = 3 units to have an area of 3 square units, the height of it must be 2. Since the parabola opens downward, the y-intercept of the equation must be -2. Filling in known values for that point, we have ...

  -2 = a(0 -1)(0 -4)

  -2 = 4a . . . . . . . . . simplify

  -2/4 = a = -1/2

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3