A quadratic polynomial maps from $[-2,3]$ onto $[0,3]$ and touches x-axis at $x=3$,then the polynomial is
- $\displaystyle \frac{3}{16}(x^{2}-6x+16)$
- $\displaystyle \frac{3}{25}(x^{2}-6x+9)$
- $\displaystyle \frac{3}{25}(x^{2}-6x+16)$
- $\displaystyle \frac{3}{16}(x^{2}-6x+9)$
Since the parabola touches the x-axis at x=3, it must be of the form a(x-3)^2. Evaluating at x=-2 gives a(-2-3)^2 = 25a. Since the range is [0,3], the maximum value at x=-2 is 3, so 25a = 3, a = 3/25. The polynomial is 3/25(x^2-6x+9).
Because the quadratic polynomial touches the x-axis at x = 3, it must have a double root at x = 3, meaning it can be written in the form f(x) = k(x - 3)^2 = kx^2 - 6kx + 9k. The function maps the interval [-2, 3] onto [0, 3], and since the vertex at x = 3 gives the minimum value of 0, the maximum value must occur at x = -2. Substituting x = -2 into the function gives k(-2 - 3)^2 = 25k, which equals the maximum value of 3. Solving 25k = 3 gives k = 3/25, so the polynomial is f(x) = (3/25)(x - 3)^2 = (3/25)(x^2 - 6x + 9).