So you would expect it to look something like this, and then asymptote towards a temperature of 27 degrees Celsius. So this is what you would expect to see, and this differential equation is consistent with that, where the rate of change, notice, this is for all t greater than zero, this is going to be a negative value, because our potato is greater than 27 for t greater than zero so this part here is going to be positive, but then you multiply positive times negative 1 4th, you're constantly going to have a negative rate of change, which makes sense, the potato is cooling down. And it also makes sense that your rate of change is proportional to the difference between the temperature of the potato and the ambient room temperature. When there's a big difference, you expect a steeper rate of change, but then when there's less of a difference, the rate of change, you can imagine, becomes less and less and less negative as we asymptote towards the ambient temperature. So with this out of the way, now let's tackle part A. Write an equation for the line tangent to the graph of h at t equals zero. Use this equation to approximate the internal temperature of the potato at time t equals three.