With "trace" still toggled on, also toggle on the "tangent" checkbox. Affirm your understanding that F'([color=#ff00ff]t[/color]) would give the slope of the displayed tangent line segment at each value of [color=#ff00ff]t[/color]. With a graph of y=F([color=#ff00ff]t[/color]) displayed via the traced points, consider what the graph of the derivative curve y=F'([color=#ff00ff]t[/color]) would look like.[br][br]How would you expect the F'([color=#ff00ff]t[/color]) graph to look?[br]How would the graphs of F'([color=#ff00ff]t[/color]) compare for the different "fixed" values of [color=#ff00ff]a[/color]=-1, [color=#ff00ff]a[/color]=0, and [color=#ff00ff]a[/color]=1?
By visually approximating the slopes of the displayed tangent line segment as you slide point [color=#ff00ff]B[/color] left/right, it appears that [color=#0000ff]f([/color][color=#ff00ff]t[/color][color=#0000ff])[/color] might give the values of those slopes F'([color=#ff00ff]t[/color]). Toggle on the "point" checkbox to see point coordinates displayed along the [color=#0000ff]f[/color] curve. The y-coordinate of the point on [color=#0000ff]f[/color] gives the slope of the line segment tangent to F.[br][br]F([color=#ff00ff]t[/color]) just vertically translates for different "fixed" values of [color=#ff00ff]a[/color], and the slopes of the tangent lines would not be affected by the vertical translations. Therefore, the F'([color=#ff00ff]t[/color]) graph would not depend on the value of [color=#ff00ff]a[/color].