Question

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.

Answer 1

Answer:

The value of a will be  $a=-\frac{1}{2}$

Step-by-step explanation:

Start by graphing the parabola and  the three points of the triangle. These points are at intersections of y=a(x-1)(x-4) and the axes

so the y = 0 points are (1,0) and (4,0).

The x = 0 intersection is when y=a(-1)(-4) or (0,4a)

The base and height of this triangle are

The base would be the distance between the y=0 intersections and the height would be the y value of the other vertex.

Hence, base=3 units and height = 4a units. Thus, area can be calculated as

$A = \frac{1}{2}\times b\times h =\frac{1}{2}\times 3\times 4a = 3$

$a=\frac{1}{2}$

∵ the parabola opens downward therefore a will be negative.

hence,  $a=-\frac{1}{2}$