The archway to the entrance of an art gallery can be modeled by y = -1/3(x-5)(x+5), where x and y are measured in feet. The x-axis represents the floor. Find the width of the arch at floor level.
The x-intercepts of your graph are -5 and 5
So what is the distance between them?
Oh, the width of the arch at floor level? That's like asking me to measure the width of a clown's shoe while it's on the run! But hey, I'm up for the challenge.
So, we have the equation y = -1/3(x-5)(x+5), right? Well, the width of the arch at floor level can be determined by finding the x-values where the arch crosses the x-axis.
To do that, we set y to zero and solve for x:
0 = -1/3(x-5)(x+5)
Now, squaring off my red nose, it seems like we have a quadratic equation on our hands. Let's bring in our math skills to solve it!
0 = -1/3(x² - 25)
0 = -1/3x² + 25/3
Now we can multiply both sides by 3 to get rid of that pesky fraction:
0 = -x² + 25
Rearranging, we have:
x² = 25
Taking the square root of both sides (and ignoring the negative root because it doesn't make sense in this context):
x = 5
So, my friend, the width of the arch at floor level is 5 feet! Just remember, even clowns can measure.
To find the width of the arch at floor level, we need to find the x-values where the arch intersects the x-axis.
Setting y = 0 in the equation y = -(1/3)(x - 5)(x + 5), we get:
0 = -(1/3)(x - 5)(x + 5)
This equation can be rewritten as:
0 = (x - 5)(x + 5)
To solve this equation, we can set each factor equal to zero:
x - 5 = 0 or x + 5 = 0
Solving for x, we have:
x = 5 or x = -5
Therefore, the arch intersects the x-axis at x = 5 and x = -5.
The width of the arch at floor level can be found as the distance between these two x-values:
Width = 5 - (-5) = 5 + 5 = 10 feet
So, the width of the arch at floor level is 10 feet.
To find the width of the arch at floor level, we need to determine the values of x that correspond to y = 0.
Given the equation y = -1/3(x-5)(x+5), we can set y equal to 0 and solve for x:
0 = -1/3(x-5)(x+5)
Since the product of two factors is equal to zero, we can use the zero product property and set each factor equal to zero:
x - 5 = 0 or x + 5 = 0
Solving these equations, we get:
x = 5 or x = -5
Therefore, the width of the arch at floor level is equal to the distance between x = 5 and x = -5, which is the absolute difference between these two values:
Width = |5 - (-5)| = 10 feet.
So, the width of the arch at floor level is 10 feet.