A quadratic polynomial maps from `[-2, 3]` onto [0, 3] and touches x-axis at x = 3, then the polynomial is

Options

A. `(3)/(16)(x^(2)-6x+16)`
B. `(3)/(25)(x^(2)-6x+9)`
C. `(3)/(25)(x^(2)-6x+16)`
D. `(3)/(16)(x^(2)-6x+9)`

Answer

To find the quadratic polynomial that maps from \([-2, 3]\) onto \([0, 3]\) and touches the x-axis at \(x = 3\), we can follow these steps:

### Step 1: Understand the properties of the polynomial
Since the polynomial touches the x-axis at \(x = 3\), it means that \(x = 3\) is a double root of the polynomial. Therefore, we can express the polynomial in the form:
\[ f(x) = a(x - 3)^2 \]
where \(a\) is a constant that determines the direction and width of the parabola.

### Step 2: Determine the range of the polynomial
The polynomial maps from \([-2, 3]\) onto \([0, 3]\). This means that the minimum value of the polynomial must be 0 (which occurs at \(x = 3\)) and the maximum value must be 3 (which occurs at the endpoint \(x = -2\)).

### Step 3: Calculate the polynomial at the endpoints
We need to find the value of \(f(-2)\):
\[ f(-2) = a(-2 - 3)^2 = a(5^2) = 25a \]
Since we want this to equal 3 (the maximum value), we set up the equation:
\[ 25a = 3 \implies a = \frac{3}{25} \]

### Step 4: Write the polynomial
Substituting \(a\) back into the polynomial, we get:
\[ f(x) = \frac{3}{25}(x - 3)^2 \]

### Step 5: Expand the polynomial
Now, we expand \(f(x)\):
\[ f(x) = \frac{3}{25}(x^2 - 6x + 9) = \frac{3}{25}x^2 - \frac{18}{25}x + \frac{27}{25} \]

### Step 6: Final polynomial
Thus, the final polynomial is:
\[ f(x) = \frac{3}{25}x^2 - \frac{18}{25}x + \frac{27}{25} \]

### Summary
The quadratic polynomial that meets the given conditions is:
\[ f(x) = \frac{3}{25}x^2 - \frac{18}{25}x + \frac{27}{25} \]