How many square units are in the area of the triangle whose vertices are the x and y intercepts of the cuure
y=(x-3)^2(x+2)?
So we are going to need those intercepts,
the x - intercepts are easy
x = 3 or x = -2
for the y-intercept, let x = 0
y = (-3)^2 (2) = 18
so the points of the triangle are (3,0), (-2,0) and (0,18)
so the base of the triangle is 5
and the height is 18
area = (1/2)(base)(height)
= (1/2)(5)(18) = 45 square units
To find the area of the triangle, we need to determine the coordinates of the vertices.
First, let's find the x-intercepts by setting y = 0:
0 = (x - 3)^2(x + 2)
By using the zero product property, we can set each factor equal to zero:
x - 3 = 0 or x + 2 = 0
Solving for x, we get:
x = 3 or x = -2
Therefore, the x-intercepts are (3, 0) and (-2, 0).
Next, let's find the y-intercept by setting x = 0:
y = (0 - 3)^2(0 + 2)
y = 9 * 2
y = 18
Therefore, the y-intercept is (0, 18).
Now, we can calculate the area of the triangle using the formula:
Area = (base * height) / 2
The base of the triangle is the distance between the x-intercepts, which is:
base = |3 - (-2)| = |5| = 5
The height of the triangle is the perpendicular distance between the y-intercept and the line determined by the x-intercepts. To find the height, we can evaluate the equation at x = 0:
y = (0 - 3)^2(0 + 2)
y = 9 * 2
y = 18
So, the height is the y-coordinate of the y-intercept, which is 18.
Now, we can calculate the area:
Area = (base * height) / 2
Area = (5 * 18) / 2
Area = 90 / 2
Area = 45 square units
Therefore, the area of the triangle whose vertices are the x and y-intercepts of the curve y = (x - 3)^2(x + 2) is 45 square units.
To find the area of the triangle with vertices at the x and y intercepts of the curve y = (x-3)^2(x+2), we will first need to find the coordinates of these intercepts.
First, let's find the x-intercepts by setting y = 0:
0 = (x-3)^2(x+2)
To solve for x, we set each factor equal to zero:
x-3 = 0 or x+2 = 0
Solving these equations, we find:
x = 3 or x = -2
So, the x-intercepts of the curve are x = 3 and x = -2.
Next, let's find the y-intercept by setting x = 0:
y = (0-3)^2(0+2)
y = 9(2)
y = 18
So, the y-intercept of the curve is y = 18.
Now that we have the coordinates of the vertices at the intercepts, let's draw the triangle on a graph.
The x-intercepts are at x = 3 and x = -2, and the y-intercept is at y = 18.
Now, let's find the lengths of the base and the height of the triangle.
The base is the distance between the x-intercepts, which is |3 - (-2)| = 5.
The height is the distance between the y-intercept and the x-axis, which is |18 - 0| = 18.
Finally, to find the area of the triangle, we use the formula:
Area = (base * height) / 2
Substituting the values we found:
Area = (5 * 18) / 2
Area = 90 / 2
Area = 45 square units
Therefore, the area of the triangle whose vertices are the x and y-intercepts of the curve y = (x-3)^2(x+2) is 45 square units.