The height of a curved support beam can be modeled by

F(x)=- x^2/288 +18
Find the height and width of the beam.



Height = units
Width = units

I bet you mean F(x)=- x^2/288 +18 x

-288 y = x^2 - 5184 x
-288 y + 6718464 = x^2 - 5184 x + 6718464
-288 y + 6718464 = (x-2592)^2
vertex at x = 2592 and height = 23328
length = 2592*2 = 5184

To find the height and width of the beam, we need to examine the equation of the curved support beam, F(x) = -x^2/288 + 18.

The height of the beam is given by the equation F(x), which represents the value of the beam at a certain position x.

To find the height, we can substitute any x-value into the equation and solve for F(x).

Let's find the height at the center of the beam, which is x = 0.

F(0) = -(0)^2/288 + 18
F(0) = 18

Therefore, the height of the beam is 18 units.

The width of the beam refers to the section of the beam where the height is nonzero. In other words, it represents the range of x-values for which the height is not zero.

To find the width, we can set the equation F(x) = 0 and solve for x.

0 = -x^2/288 + 18

Rearranging the equation:

x^2/288 = 18

Multiplying both sides by 288:

x^2 = 5184

Taking the square root of both sides:

x = ±√5184

x ≈ ± 72

Therefore, the width of the beam is 144 units (2 * 72).

Height = 18 units
Width = 144 units

To find the height and width of the beam, we need to understand the variables and equations involved.

In this case, the height of the curved support beam is represented by the equation F(x) = -x^2/288 + 18.

To find the height, we need the value of F(x) when x is equal to zero because the height of the beam is measured at the x-axis.

Substituting x = 0 into the equation, we have:
F(0) = -(0^2)/288 + 18 = 18

Therefore, the height of the beam is 18 units.

To find the width of the beam, we need to determine the x-values where F(x) is equal to zero. The width corresponds to the interval in which the beam intersects or touches the x-axis.

Setting F(x) equal to zero, we have:
- x^2/288 + 18 = 0

Multiplying both sides of the equation by -288, we get:
x^2 = 18 * 288

Taking the square root of both sides, we have:
x = ± sqrt(18 * 288)

Simplifying further, we have:
x ≈ ± 24

Therefore, the width of the beam is approximately 24 units.