+87 How many square units are in the area of the triangle whose vertices are the x- and y- intercepts of the function f(x)=(x−1)^2(x+3)
+539 I just graphed it
But you can also find the values of the x and y-intercepts
To find the y-intercept, x has to be 0, 0-1 is -1, and -1^2 is 1. 0+3 is 3, and 3*1 is 3. So the y-intercept would be (0,3)
To find the x-intercepts, y has to be zero. So either (x+3) or (x-1)^2 should be 0.
For (x+3), to make it 0, x has to be -3.
Since squaring any number except 0 can't get you zero, the square doesn't matter. So x-1 is 0. x=1. So our x-intercepts are (-3,0) and (1,0).
The base has length 4 and the triangle has a height of 3. (4*3)/2 is 6.
+539 I just graphed it
But you can also find the values of the x and y-intercepts
To find the y-intercept, x has to be 0, 0-1 is -1, and -1^2 is 1. 0+3 is 3, and 3*1 is 3. So the y-intercept would be (0,3)
To find the x-intercepts, y has to be zero. So either (x+3) or (x-1)^2 should be 0.
For (x+3), to make it 0, x has to be -3.
Since squaring any number except 0 can't get you zero, the square doesn't matter. So x-1 is 0. x=1. So our x-intercepts are (-3,0) and (1,0).
The base has length 4 and the triangle has a height of 3. (4*3)/2 is 6.
