Determine the value of the 5-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.

To find the value of the \( y \)-intercept of a line with a given slope that is tangent to the curve \( y = -2x^2 + 5x - 4 \), we need to follow these steps:

### Step 1: Find the derivative of the curve
Since the slope of the tangent line to the curve at any point is given by the derivative of the curve, we first find the derivative of the function:
\[
y' = \frac{d}{dx}(-2x^2 + 5x - 4) = -4x + 5.
\]

### Step 2: Set the derivative equal to the slope of the tangent line
The slope of the line is given as \( 1 \). To find the point(s) on the curve where the slope of the tangent is 1, we set the derivative equal to 1:
\[
-4x + 5 = 1.
\]

### Step 3: Solve for \( x \)
\[
-4x + 5 = 1,
\]
\[
-4x = 1 - 5,
\]
\[
-4x = -4,
\]
\[
x = 1.
\]

### Step 4: Find the corresponding \( y \)-coordinate on the curve
Substitute \( x = 1 \) back into the original curve equation to find the \( y \)-coordinate:
\[
y = -2(1)^2 + 5(1) - 4,
\]
\

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.

This detailed solution explains how to find the y-intercept of a line with a given slope that is tangent to the curve y = -2x^2 + 5x - 4. The process involves finding the derivative of the quadratic curve, setting it equal to the slope of the line, and solving for the point of tangency. Ideal for students in Grades 11-12 learning about calculus and derivatives.

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