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 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)

 Jan 5, 2021

Best Answer 

 #1
avatar+487 
+1

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.

 Jan 5, 2021
 #1
avatar+487 
+1
Best Answer

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.

MooMooooMooM Jan 5, 2021

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