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We are given the curve equation:

\[
y = \frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x + 5
\]

We need to find the points on the curve where the gradient of the tangent is 4. The gradient of the tangent to the curve at any point is given by the derivative \( \frac{dy}{dx} \) of the curve equation.

### Step 1: Differentiate the curve equation
Let's compute \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{3}x^3 - \frac{1}{2}x^2 + 2x + 5 \right)
\]

Using the power rule, the derivative becomes:

\[
\frac{dy}{dx} = x^2 - x + 2
\]

### Step 2: Set the derivative equal to 4
Since we are asked to find the points where the gradient is 4, we set \( \frac{dy}{dx} = 4 \):

\[
x^2 - x + 2 = 4
\]

### Step 3: Solve for \( x \)
Now solve the quadratic equation:

\[
x^2 - x + 2 - 4 = 0
\]
\[
x^2 - x - 2 = 0


A curve has the equation y = (1/3)x^3 - (1/2)x^2 + 2x + 5. Find the coordinates of the points on the curve where the tangent has gradient 4.

This problem involves using differentiation to find the points on the curve y = (1/3)x^3 - (1/2)x^2 + 2x + 5 where the gradient of the tangent is 4. The solution includes finding the derivative of the cubic equation and solving a quadratic equation to determine the x-values. Suitable for students in Grades 10-12.

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