Do what is in the parentheses then add
Answer:
<h2>y = 5x + 11</h2>
Step-by-step explanation:


Answer:
- zeros: x = -3, -1, +2.
- end behavior: as x approaches -∞, f(x) approaches -∞.
Step-by-step explanation:
I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.
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The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...
x² +x -6
which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.
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For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).
Answer:
1
Step-by-step explanation:
1 to the power of anything is still one
1 * 1 * 1 * 1 = 1