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Determine the value of ' \( k \) ' if the line given by \( y=-2 x+k \) is a by tangent to the function given by \( y=-x^{2}+4 x+5 \)

Ask by Henry Bowman. in South Africa
Jan 23,2025

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Answer

\( k = 14 \)

Solution

To determine the value of \( k \) such that the line \( y = -2x + k \) is tangent to the parabola \( y = -x^2 + 4x + 5 \), follow these steps: 1. **Set the Equations Equal:** For the line to be tangent to the parabola, they must intersect at exactly one point. Set the two equations equal to each other: \[ -2x + k = -x^2 + 4x + 5 \] 2. **Rearrange the Equation:** Bring all terms to one side to form a quadratic equation: \[ x^2 - 6x + (k - 5) = 0 \] 3. **Condition for Tangency:** A quadratic equation \( ax^2 + bx + c = 0 \) has exactly one real solution (i.e., the line is tangent to the parabola) if its discriminant \( D \) is zero. The discriminant is given by: \[ D = b^2 - 4ac \] For the quadratic \( x^2 - 6x + (k - 5) = 0 \): \[ a = 1,\quad b = -6,\quad c = k - 5 \] So, \[ D = (-6)^2 - 4(1)(k - 5) = 36 - 4(k - 5) \] 4. **Solve for \( k \):** Set the discriminant to zero: \[ 36 - 4(k - 5) = 0 \] \[ 36 - 4k + 20 = 0 \] \[ 56 - 4k = 0 \] \[ 4k = 56 \] \[ k = 14 \] **Answer:** \( k = 14 \)

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To find the value of \( k \) where the line \( y = -2x + k \) is a tangent to the parabola \( y = -x^2 + 4x + 5 \), we first set the equations equal to each other: \[ -2x + k = -x^2 + 4x + 5 \] Rearranging gives us: \[ x^2 + 6x + (5 - k) = 0 \] For the line to be tangent to the parabola, the quadratic equation must have exactly one solution, which occurs when the discriminant is zero. The discriminant \( D \) is given by: \[ D = b^2 - 4ac \] Here, \( a = 1 \), \( b = 6 \), and \( c = 5 - k \): \[ D = 6^2 - 4(1)(5 - k) = 36 - 20 + 4k = 16 + 4k \] Setting the discriminant \( D \) equal to zero: \[ 16 + 4k = 0 \] Solving for \( k \): \[ 4k = -16 \quad \Rightarrow \quad k = -4 \] Thus, the value of \( k \) is \( -4 \).

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