Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. \( h(t)=-2 t^{3},(-2,16) \)

To find the slope of the function \( h(t) = -2t^3 \) at the point \((-2, 16)\) and to determine the equation of the tangent line at that point, follow these steps: ### 1. Verify the Point on the Function First, ensure that the point \((-2, 16)\) lies on the graph of \( h(t) \). \[ h(-2) = -2(-2)^3 = -2(-8) = 16 \] Since \( h(-2) = 16 \), the point \((-2, 16)\) is indeed on the graph. ### 2. Find the Derivative (Slope Function) The slope of the tangent line at any point on the function is given by the derivative of \( h(t) \). \[ h(t) = -2t^3 \] \[ h'(t) = \frac{d}{dt}(-2t^3) = -6t^2 \] ### 3. Calculate the Slope at \( t = -2 \) Substitute \( t = -2 \) into the derivative to find the slope at that point. \[ h'(-2) = -6(-2)^2 = -6 \times 4 = -24 \] So, the slope of the tangent line at \((-2, 16)\) is **-24**. ### 4. Find the Equation of the Tangent Line Using the point-slope form of a line: \[ y - y_1 = m(t - t_1) \] Where: - \( m = -24 \) (slope) - \( (t_1, y_1) = (-2, 16) \) Substitute these values into the equation: \[ y - 16 = -24(t - (-2)) \] \[ y - 16 = -24(t + 2) \] \[ y = -24t - 48 + 16 \] \[ y = -24t - 32 \] ### **Final Answer** - The slope at (–2, 16) is –24. - An equation for the tangent line is y = –24 t – 32.