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

Options

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

Answer

To solve the problem step-by-step, we need to find a quadratic polynomial \( f(x) = ax^2 + bx + c \) that meets the given conditions.

### Step 1: Understand the properties of the polynomial
The polynomial maps from the interval \([-2, 3]\) onto \([0, 3]\) and touches the x-axis at \(x = 3\). This means that \(f(3) = 0\) and \(x = 3\) is a double root of the polynomial.

### Step 2: Formulate the polynomial
Since \(x = 3\) is a double root, we can express the polynomial in the form:
\[ f(x) = a(x - 3)^2 \]
where \(a\) is a constant that we need to determine.

### Step 3: Expand the polynomial
Expanding the polynomial gives:
\[ f(x) = a(x^2 - 6x + 9) = ax^2 - 6ax + 9a \]
Thus, we have:
- \(b = -6a\)
- \(c = 9a\)

### Step 4: Determine the range of the polynomial
Since the polynomial maps from \([-2, 3]\) onto \([0, 3]\), we need to find the values of \(f(-2)\) and \(f(3)\):
- \(f(3) = 0\) (as it touches the x-axis)
- We need to find \(f(-2)\):
\[ f(-2) = a((-2) - 3)^2 = a(25) = 25a \]
We want \(f(-2) = 3\) (the maximum value in the range).

### Step 5: Set up the equation
Setting \(25a = 3\):
\[ a = \frac{3}{25} \]

### Step 6: Find \(b\) and \(c\)
Using \(a = \frac{3}{25}\):
- \(b = -6a = -6 \times \frac{3}{25} = -\frac{18}{25}\)
- \(c = 9a = 9 \times \frac{3}{25} = \frac{27}{25}\)

### Step 7: Write the polynomial
Now we can write the polynomial:
\[ f(x) = \frac{3}{25}x^2 - \frac{18}{25}x + \frac{27}{25} \]

### Step 8: Check the options
We need to check which option corresponds to our polynomial:
- The polynomial can be rewritten as:
\[ f(x) = \frac{3}{25}(x^2 - 6x + 9) = \frac{3}{25}(x - 3)^2 \]
This matches option (b) from the provided options.

### Final Answer
The polynomial is:
\[ f(x) = \frac{3}{25}x^2 - \frac{18}{25}x + \frac{27}{25} \]
Thus, the correct option is (b).