Least Degree Polynomial Cae158

1. **State the problem:** We need to find the least possible degree of a polynomial whose graph is a straight line crossing the y-axis at $-1$ and the x-axis at $2$. 2. **Understand the graph:** The graph is a straight increasing line. A polynomial that graphs as a straight line is a polynomial of degree $1$ (a linear polynomial). 3. **Recall polynomial degree rules:** - Degree $0$ polynomial is a constant function (horizontal line). - Degree $1$ polynomial is a linear function (straight line with non-zero slope). - Higher degree polynomials have curves, turning points, or more complex shapes. 4. **Check the given points:** - The line crosses the y-axis at $-1$, so when $x=0$, $y=-1$. - The line crosses the x-axis at $2$, so when $x=2$, $y=0$. 5. **Form the linear polynomial:** Let the polynomial be $p(x) = mx + b$. Using the y-intercept: $p(0) = b = -1$. Using the x-intercept: $p(2) = 2m + b = 0$. Substitute $b = -1$: $$2m - 1 = 0$$ $$2m = 1$$ $$m = \frac{1}{2}$$ 6. **Final polynomial:** $$p(x) = \frac{1}{2}x - 1$$ 7. **Conclusion:** The polynomial is degree $1$, which is the least possible degree for the given graph. **Final answer:** The least possible degree of the polynomial is $1$.