Determine the value of the y-intercept of a line with the given slope that is a tangent line to the given curve. y = -2x^2 + 5x _ 4 and a line with a slope of 1. First find the b value in y = 1x +b

We are tasked with finding the equation of the tangent line to the curve \( y = -2x^2 + 5x - 4 \) that has a slope of 1. Specifically, we need to find the value of \( b \) in the line equation \( y = x + b \), which is the y-intercept.

### Steps to solve:

1. **Find the derivative of the curve equation**:
   The derivative will give us the slope of the tangent line at any point \( x \). The curve is \( y = -2x^2 + 5x - 4 \).
   
   The derivative \( \frac{dy}{dx} \) is:
   \[
   \frac{dy}{dx} = \frac{d}{dx}(-2x^2 + 5x - 4) = -4x + 5
   \]
   
2. **Set the derivative equal to 1 (the slope of the tangent line)**:
   To find the point where the slope of the tangent line is 1, we set the derivative equal to 1:
   \[
   -4x + 5 = 1
   \]
   Solve for \( x \):
   \[
   -4x = 1 - 5 = -4
   \]
   \[
   x = 1
   \]

3. **Find the y-coordinate of the point of tangency**:
   Now that we know \( x = 1 \), we substitute it into the original curve equation to find the corresponding \( y \)-coordinate:
   \[
   y = -2(1)^2 + 5(1) - 4 = -2 + 5 - 4 = -1
   \]
   So the point of tangency is \( (1, -1) \).

4. **Use the point-slope form of the equation of a line**:
   The slope of the tangent line is 1, and the point of tangency is \( (1, -1) \). The equation of the line is:
   \[
   y - (-1) = 1(x - 1)
   \]
   Simplifying:
   \[
   y + 1 = x - 1
   \]
   \[
   y = x - 2
   \]
   So the equation of the tangent line is \( y = x - 2 \).

5. **Find the value of \( b \)**:
   The equation of the tangent line is in the form \( y = x + b \), so comparing \( y = x - 2 \) with \( y = x + b \), we find that:
   \[
   b = -2
   \]

Thus, the value of the y-intercept \( b \) is \( -2 \).

Would you like more details or have any other questions? Here are some related questions to further your understanding:

1. What does the derivative of a function represent in terms of the graph?
2. How do we find the slope of a tangent line to a curve at a specific point?
3. Can the tangent line at any point have a negative slope? How would we find that?
4. What are the different forms of the equation of a line?
5. How does the second derivative of a function relate to the concavity of the graph?

**Tip:** When solving problems involving tangent lines, always begin by finding the derivative of the curve to identify where the slopes match.

Determine the value of the y-intercept of a line with the given slope that is a tangent line to the given curve. y = -2x^2 + 5x - 4 and a line with a slope of 1. First find the b value in y = 1x + b.

Discover how to find the y-intercept of a tangent line to the curve y = -2x^2 + 5x - 4 with a slope of 1. This detailed solution includes derivative calculations and the use of point-slope form. Suitable for students in Grades 10-12.

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