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# Can anyone help me solve this Algebra 2 Problem, i literally cant figure it out

Accepted Solution

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Answer:Zeros : 1 , -1, 3Degree : 4End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞Y - intercept : -3Extra Points: (0,-3), (2,-3)Step-by-step explanation:f(x) = 0 to find the zeros$$Therefore (x+1)(x-1)^{2} (x-3) = 0$$Clearly x = -1,1,3Here 1 is a repeating root as it is (x-1)²Degree is highest power of x in f(x)Clearly it is x*x²*x = x⁴ is the maximum power of xThus degree is 4Looking at end behavior we substitute x->∞ and x-> -∞Clearly f(x)>0 as all terms are positive and f(x)->∞Similarly when x->-∞ f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞Y-Intercept is f(0)f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3Thus Y-Intercept is -3Substitute x = 0 , 2 for extra points Thus f(0) = -3 and f(2) = -3Thus points on the graph (0,-3), (0,2)We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.