A curve is defined by the equation y = x^2 - 6x + 11. What is the minimum value of y on th… — JAMB UTME

A curve is defined by the equation $y = x^2 - 6x + 11$ . What is the minimum value of $y$ on this curve?

Answer: 2

The minimum value of

y

on the curve is 2.

Differentiating

y = x^2 - 6x + 11

gives

\frac{dy}{dx} = 2x - 6

. Setting this derivative to zero yields

2x - 6 = 0

, so

x = 3

. Substituting

x = 3

into the original function gives

y = (3)^2 - 6(3) + 11 = 2

. Since

\frac{d^2y}{dx^2} = 2 > 0

, the point at

x = 3

is a local minimum, making 2 the minimum value of

y

Find the first derivative of the curve function.

\frac{dy}{dx} = 2x - 6

Stationary points occur where the derivative is equal to zero.

Solve for the

x

-coordinate at the stationary point.

2x - 6 = 0 \implies x = 3

Setting the derivative to zero determines the input value where the slope is horizontal.

Calculate the corresponding

y

-value at

x = 3

y = (3)^2 - 6(3) + 11 = 2

Evaluating the original equation at

x = 3

yields the minimum value of

y

Finding the minimum value of a function using differentiation

Question