I'm pretty sure I have to factor to get the x and y intercepts here but I have no idea how to do so. I'm supposed to find them without using guess and check for all x intercepts. Can anyone help me with this question?
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f(0) = -24 ---> (0,-24)
f(1) = 0 ---> x-1 is a factor ---> f(x) = -2(x-1)(x^2+x-12) = 0 ---> x = 1 , -4 , 3
f(1) = 0 ---> x-1 is a factor ---> f(x) = -2(x-1)(x^2+x-12) = 0 ---> x = 1 , -4 , 3
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The y-intercept is easy. Y-intercepts are just where the curve cross the y-axis... Since these points all have an x-value of 0 (every point on the y-axis has an x-value of 0), we plug 0 in for x and we get the y-intercept is -24.
For the x-intercept, we set the function equal to 0 and solve for x.
-2x^3 + 26x - 24 = 0
You're right. We need to factor. It would be best to pull a -2 out of this thing first.
-2(x^2 - 13x + 12) = 0
Then, we can factor this guy.
-2(x - 12)(x - 1) = 0
So, our solutions are x = 12 and x = 1.
So x-intercepts are at 12 and 1. Y-intercept is at -24.
For the x-intercept, we set the function equal to 0 and solve for x.
-2x^3 + 26x - 24 = 0
You're right. We need to factor. It would be best to pull a -2 out of this thing first.
-2(x^2 - 13x + 12) = 0
Then, we can factor this guy.
-2(x - 12)(x - 1) = 0
So, our solutions are x = 12 and x = 1.
So x-intercepts are at 12 and 1. Y-intercept is at -24.
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x intercepts : y = 0 ,
-2x^3 + 26x - 24 = 0 ,
(x-1)(x-2)(x+3) = 0
x = 1 , x = 3 , x = -4
y intercepts : x = 0 , y = -24
-2x^3 + 26x - 24 = 0 ,
(x-1)(x-2)(x+3) = 0
x = 1 , x = 3 , x = -4
y intercepts : x = 0 , y = -24