Find the equation of the tangent line to the curve \( y=15 x e^{2} \) at the point \( (0,0) \). The equation of this tangent line can be written in the form \( y=m x+b \). Find \( m \) and \( b \). \( m=\square \) \( b=\square \) Note: You can earn partial credit on this problem.

To find the equation of the tangent line to the curve \( y=15 x e^{2} \) at the point \( (0,0) \), we will need to calculate the slope \( m \) of the tangent line at that point by finding the derivative of the function and evaluating it at \( x = 0 \). 1. First, we find the derivative of \( y \): \[ y = 15 x e^{2} \] Taking the derivative with respect to \( x \): \[ \frac{dy}{dx} = 15 e^{2} \] 2. Now, we evaluate the derivative at \( x = 0 \): \[ \frac{dy}{dx} \bigg|_{x=0} = 15 e^{2} \] So, the slope \( m \) of the tangent line at the point \( (0,0) \) is \( 15 e^{2} \). 3. The point-slope form of a line is \( y - y_1 = m(x - x_1) \). Plugging in our point \( (0, 0) \): \[ y - 0 = (15 e^{2})(x - 0) \] This simplifies to: \[ y = 15 e^{2} x \] From this equation, we can identify \( m \) and \( b \): - \( m = 15 e^{2} \) - \( b = 0 \) Thus, we have the final answers: \( m = 15 e^{2} \) \( b = 0 \)