Question

If the tangent to the curve \( y = ax^2 + bx + c \) at the point \( (1,1) \) is parallel to the x-axis, and the curve passes through the point \( (2,3) \), then find the value of \( a + b + c \):

• A. 1
• B. 3
• C. 4
• D. 5

Hint:

When the tangent is horizontal, set the derivative equal to zero at the given point. Use the point conditions to find unknown coefficients. Always verify your answers with all given conditions.

Answer 1

The Correct Option is A

Given the curve: \[ y = ax^2 + bx + c \] 1) Since the tangent at \( (1,1) \) is parallel to the x-axis, the slope at \( x=1 \) must be zero. The slope is given by the derivative: \[ \frac{dy}{dx} = 2ax + b \] At \( x=1 \): \[ 0 = 2a(1) + b = 2a + b \implies b = -2a \] 2) The point \( (1,1) \) lies on the curve: \[ 1 = a(1)^2 + b(1) + c = a + b + c \] 3) The point \( (2,3) \) lies on the curve: \[ 3 = a(2)^2 + b(2) + c = 4a + 2b + c \] Substitute \( b = -2a \) into the above two equations: From 2): \[ 1 = a + (-2a) + c = -a + c \implies c = 1 + a \] From 3): \[ 3 = 4a + 2(-2a) + c = 4a - 4a + c = c \] Therefore, \[ c = 3 \] From the earlier expression for \( c \): \[ c = 1 + a = 3 \implies a = 2 \] Then, \[ b = -2a = -4 \] Finally, calculate \( a + b + c \): \[ a + b + c = 2 + (-4) + 3 = 1 \]