Question

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

Answer 1

Zeros : 1 , -1, 3

Degree : 4

End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞

Y - intercept : -3

Extra Points: (0,-3), (2,-3)

Explanation:

f(x) = 0 to find the zeros

`Therefore (x+1)(x-1)^(2) (x-3) = 0`

Clearly x = -1,1,3

Here 1 is a repeating root as it is (x-1)²

Degree is highest power of x in f(x)

Clearly it is x*x²*x = x⁴ is the maximum power of x

Thus degree is 4

Looking 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 = -3

Thus Y-Intercept is -3

Substitute x = 0 , 2 for extra points

Thus f(0) = -3

and f(2) = -3

Thus 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.