The equation of a curve is y=2x squared -6x+3. Find dy/dx and hence find the coordinates of the point on the curve at which the tangent is parallel to the line y=2x+4

To solve this, let's break it down into two parts:

### 1. **Find the derivative \( \frac{dy}{dx} \):**

The equation of the curve is:

\[
y = 2x^2 - 6x + 3
\]

To find \( \frac{dy}{dx} \), we'll differentiate \( y \) with respect to \( x \).

\[
\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 6x + 3)
\]

Using the power rule:

\[
\frac{dy}{dx} = 4x - 6
\]

### 2. **Find the point where the tangent is parallel to the line \( y = 2x + 4 \):**

The slope of the tangent to the curve is given by \( \frac{dy}{dx} \), and we need to find where this slope is equal to the slope of the line \( y = 2x + 4 \).

For a line in the form \( y = mx + c \), the slope is \( m \). Here, the slope is \( 2 \), as \( y = 2x + 4 \).

So, we set \( \frac{dy}{dx} = 2 \):

\[
4x - 6 = 2
\]

Solving for \( x \):

\[
4x = 2 + 6
\]
\[
4x = 8
\]
\[
x = 2
\]

### 3. **Find the corresponding \( y \)-coordinate:**

Substitute \( x = 2 \) back into the original equation of the curve to find \( y \):

\[
y = 2(2)^2 - 6(2) + 3
\]
\[
y = 2(4) - 12 + 3
\]
\[
y = 8 - 12 + 3 = -1
\]

Thus, the coordinates of the point where the tangent is parallel to the line are \( (2, -1) \).

### Final Answer:
The coordinates of the point are \( (2, -1) \).

Would you like more details or have any questions? Here are five related questions to expand your understanding:

1. How would you differentiate a cubic function?
2. What does it mean geometrically for a tangent to be parallel to a line?
3. Can you find the equation of the tangent line at the point \( (2, -1) \)?
4. How would the problem change if the slope of the given line were different?
5. What would the second derivative \( \frac{d^2y}{dx^2} \) tell us about the curve?

**Tip:** Remember that parallel lines always have the same slope, which helps when comparing functions!

The equation of a curve is y=2x^2 - 6x + 3. Find dy/dx and hence find the coordinates of the point on the curve at which the tangent is parallel to the line y=2x+4.

Solve a problem involving differentiation of the curve y=2x^2 - 6x + 3 to find dy/dx. Learn how to determine the point where the tangent to the curve is parallel to the line y=2x+4. This step-by-step solution explores key concepts such as the power rule for differentiation and the slope of parallel lines. Suitable for students in Grades 10-12.

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