If the curve `y=a x^2-6x+b` pass through `(0,2)` and has its tangent parallel to the x-axis at `x=3/2,` then find the values of `aa n dbdot`

Answer

To solve the problem step by step, we will follow the given conditions about the curve and use derivatives to find the values of \( a \) and \( b \).

### Step 1: Write the equation of the curve
The curve is given by the equation:
\[ y = ax^2 - 6x + b \]

### Step 2: Use the point (0, 2)
Since the curve passes through the point \((0, 2)\), we can substitute \( x = 0 \) and \( y = 2 \) into the equation:
\[ 2 = a(0)^2 - 6(0) + b \]
This simplifies to:
\[ 2 = b \]
Thus, we find:
\[ b = 2 \]

### Step 3: Find the derivative of the curve
Next, we need to find the derivative of the curve to determine the slope of the tangent line:
\[ \frac{dy}{dx} = \frac{d}{dx}(ax^2 - 6x + b) = 2ax - 6 \]

### Step 4: Set the derivative to zero for the tangent parallel to the x-axis
We know that the tangent is parallel to the x-axis at \( x = \frac{3}{2} \). Therefore, we set the derivative equal to zero:
\[ 0 = 2a\left(\frac{3}{2}\right) - 6 \]
This simplifies to:
\[ 0 = 3a - 6 \]

### Step 5: Solve for \( a \)
Rearranging the equation gives:
\[ 3a = 6 \implies a = \frac{6}{3} = 2 \]

### Conclusion
We have found the values of \( a \) and \( b \):
\[ a = 2, \quad b = 2 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ \boxed{2} \text{ and } \boxed{2} \]