The zeros and turning point of y=4x^2-8x-45 are vertices of a triangle. Find the area of this triangle?

Answer 1

Find the zeroes:
0 = 4x^2 - 8x - 45
0 = x^2 - 2x - 45/4
x = [2 ± √(4 - 4(1)(-45/4)) ] / 2
x = [2 ± √(49)) ] / 2
x = (2+7)/2, (2-7)/2
x = 9/2, -5/2
So the base of this triangle has a length of 9/2 - (-5/2) = 14/2 = 7.

The axis of symmetry would be right in the middle of these points, so it's at x = (9/2 + -5/2) / 2 = 1. NOT -1 as the user before be stated. Since f(1) = -49, the vertex point is at (1, - 49). This means the height of the triangle must be 49, since the base is at y=0.

As with any triangle, take half of the base times the height to find the area: (1/2)(7)(49) = 343/2

Answer 2

The first thing we need to do is find the zeros and "turning point". Well,
4x^2 - 8x - 45
=(2x+5) (2x-9)

So the zeros are (-2.5, 0) and (4.5, 0).

Now to find the inflection ("turning") point, we take the derivative and set it equal to zero:
8x - 8
=8(x-1)=0

So that (1, -49) is our inflection point.

Now there are several ways to find the area of the triangle. In this case, the well-known 1/2 * b * h is the best choice, taking the segment on the x-axis as the base:
A=1/2 * |4.5 - (-2.5)| * |-49 - 0|
=1/2 * 7 * 49
=171.5

EDIT: Wow, four "top contributors" in under 12 minutes...nicely done

Answer 3

oops, Sahsjing got the sign of the turning point wrong.

With zeros at 4.5 and -2.5, line of symmetry is halfway between at (4.5 - 2.5)/2 = 2/2 = +1.

Then vertex is at x = 1, y = 4(1)² - 8(1) - 45 = 4 - 8 - 45 = -49, and

base of triangle is 4.5 - (-2.5) = 7, and height is 49. area is half of base times height.

Answer 4

y = (2x-9)(2x+5)
Zeros: x = 4.5, -2.5
Turning point:
x = (4.5-2.5)/2 = 1
f(1) = -49

A = (1/2)(4.5+2.5)(49) = 171.5