Questions: Consider the function which has intercepts (-3,0), (6,0) and (0,2). What is the polynomial of least degree that goes through these points? Quartic Linear Quadratic Cubic

Solution Steps

To find the polynomial of least degree that goes through the given points \((-3,0)\), \((6,0)\), and \((0,2)\), we need to determine the polynomial that fits these intercepts. Since the polynomial has two x-intercepts and one y-intercept, it must be at least a quadratic polynomial. We can use the factored form of a quadratic polynomial and then determine the coefficients by substituting the given points.

1. Start with the general form of a quadratic polynomial: \( y = a(x + 3)(x - 6) \).
2. Use the y-intercept \((0,2)\) to find the coefficient \(a\).
3. Substitute \(x = 0\) and \(y = 2\) into the polynomial to solve for \(a\).

We start with the general form of a quadratic polynomial that has x-intercepts at \((-3, 0)\) and \((6, 0)\): \[ y = a(x + 3)(x - 6) \]

To find the coefficient \(a\), we use the y-intercept \((0, 2)\). Substituting \(x = 0\) and \(y = 2\) into the polynomial gives: \[ 2 = a(0 + 3)(0 - 6) \] This simplifies to: \[ 2 = -18a \]

Rearranging the equation, we find: \[ a = -\frac{1}{9} \]

Substituting \(a\) back into the polynomial, we have: \[ y = -\frac{1}{9}(x + 3)(x - 6) \]

Expanding the polynomial results in: \[ y = -\frac{1}{9}(x^2 - 3x - 18) \] This simplifies to: \[ y = -\frac{1}{9}x^2 + \frac{1}{3}x + 2 \]

Final Answer