Gradient function of the curve is y=6x+8. Find equation of the curve passing through (1,2)

To find the equation of the curve, we need to integrate the gradient function \( \frac{dy}{dx} = 6x + 8 \) since the gradient of the curve is the derivative of the curve's equation.

### Step 1: Integrate the gradient function
We integrate the gradient function with respect to \( x \):

\[
y = \int (6x + 8) \, dx
\]

### Step 2: Perform the integration
\[
y = \int 6x \, dx + \int 8 \, dx
\]
\[
y = 3x^2 + 8x + C
\]

where \( C \) is the constant of integration.

### Step 3: Use the given point \((1, 2)\) to find \( C \)
We know the curve passes through the point \( (1, 2) \). Substitute \( x = 1 \) and \( y = 2 \) into the equation:

\[
2 = 3(1)^2 + 8(1) + C
\]
\[
2 = 3 + 8 + C
\]
\[
2 = 11 + C
\]
\[
C = 2 - 11 = -9
\]

### Step 4: Write the equation of the curve
Now that we have \( C = -9 \), the equation of the curve is:

\[
y = 3x^2 + 8x - 9
\]

This is the equation of the curve passing through the point \( (1, 2) \).

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you find the tangent line to the curve at a specific point?
2. What is the geometric significance of the constant of integration \( C \)?
3. How can you determine the points where the curve intersects the x-axis?
4. What happens to the curve's equation if the gradient function changes?
5. Can you find the area under the curve \( y = 3x^2 + 8x - 9 \) between two points?

**Tip:** When integrating a function, always remember to include the constant of integration \( C \). This constant is determined by initial conditions or specific points on the curve.

Discover how to find the equation of a curve from its gradient function y = 6x + 8, with a detailed step-by-step solution. This problem involves key calculus concepts, including integration and curve equations, and is ideal for students in Grades 10-12.

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