An even function has 3 distinct roots and end behaviours where f(x) -> - infinity as x-> +- infinity. State one possible equation for this even function, and justify your choices.
Any ideas on any of these?
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almost -- except you got the end-behavior sign wrong.
and your p(x) only has two distinct real roots, since (x^2+1) has no real roots.
Try p(x) = -x^2 (x-1)(x+1)
To find an equation for the even function that satisfies the given conditions, we can start by analyzing the requirements.
1. The function has three distinct roots: This means that the function crosses the x-axis at three different points. Let's denote these roots as x1, x2, and x3.
2. The end behavior as x approaches positive or negative infinity is f(x) -> -infinity: This indicates that the function approaches negative infinity as x becomes extremely large or extremely small.
Based on these conditions, one possible equation for the even function could be a polynomial function with a degree greater than or equal to six (to account for three distinct roots) and leading coefficient negative. Let's consider a polynomial equation of degree six for simplicity:
f(x) = a(x - x1)(x - x2)(x - x3)(x - r1)(x - r2)(x - r3)
Here, the variables r1, r2, and r3 are any additional roots that the function may have. Since we want the function to approach negative infinity as x goes to infinity, the leading coefficient 'a' must be negative. The choice of specific values for x1, x2, x3, r1, r2, and r3 depends on the desired roots and additional properties of the function.
Please note that there can be multiple valid equations satisfying these conditions. The given equation provides a general structure that meets the specified criteria.