In the xy-plane , a line that has the equation y=c for some constant c intersects a parabola at exactly one point. If the parabola has the equation `y=-x^2 + 5x`, what is the value of c ?

Explanation

To find the value of \( c \) such that the line \( y = c \) intersects the parabola \( y = -x^2 + 5x \) at exactly one point, we can follow these steps:

### Step 1: Set the equations equal to each other
We start by substituting \( y = c \) into the equation of the parabola:

\[ c = -x^2 + 5x \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ x^2 - 5x + c = 0 \]

### Step 3: Identify the coefficients
This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -5 \)
- \( c = c \) (the constant we are trying to find)

### Step 4: Use the discriminant
For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Substituting our values into the discriminant formula:

\[ D = (-5)^2 - 4(1)(c) = 25 - 4c \]

### Step 5: Set the discriminant to zero
Since we want the line to intersect the parabola at exactly one point, we set the discriminant equal to zero:

\[ 25 - 4c = 0 \]

### Step 6: Solve for \( c \)
Now, we solve for \( c \):

\[ 25 = 4c \]

\[ c = \frac{25}{4} \]

### Conclusion
Thus, the value of \( c \) is:

\[ \boxed{\frac{25}{4}} \]

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