Does the parabola \( y=3 x^{2}-25 x+8 \) have a tangent line whose slope is -1 ? If so, find an equation for the line and the point of tangency. If not, why not?

To determine if the parabola \( y=3x^{2}-25x+8 \) has a tangent line with a slope of -1, we first find the derivative of the function to get the slope of the tangent line at any point on the parabola. The derivative \( y' \) is calculated as follows: \[ y' = \frac{d}{dx}(3x^2 - 25x + 8) = 6x - 25. \] Next, we set the derivative equal to -1 to find the x-coordinate where the tangent line has that slope: \[ 6x - 25 = -1. \] Solving for \( x \): \[ 6x = 24 \quad \Rightarrow \quad x = 4. \] Now we substitute \( x=4 \) back into the original equation to find the corresponding y-coordinate: \[ y = 3(4^2) - 25(4) + 8 = 3(16) - 100 + 8 = 48 - 100 + 8 = -44. \] So the point of tangency is \( (4, -44) \). Now, we can find the equation of the tangent line using the point-slope form of a line: \[ y - y_1 = m(x - x_1), \] where \( m = -1 \), \( (x_1, y_1) = (4, -44) \): \[ y + 44 = -1(x - 4) \quad \Rightarrow \quad y + 44 = -x + 4 \quad \Rightarrow \quad y = -x - 40. \] Thus, the equation of the tangent line with a slope of -1 is \( y = -x - 40 \). Yes, the parabola does have a tangent line with a slope of -1 at the point \( (4, -44) \).