Which transformations could have been used to create A'B'C'? Choose all that apply.
1 answer:
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Answer:
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)
Step-by-step explanation:
f(x) = 0 to find the zeros
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.