Final answer:
To draw the tangent to the function f(x) = x^2 + 3 at the point (1, f(1)), calculate the function's derivative to find the slope, use the point-slope equation to find the tangent line's equation, and plot another point on this line, such as (1.1, 4.2).
Step-by-step explanation:
To draw the tangent line to the function f(x) = x^2 + 3 at the point (1, f(1)), we first need to calculate the derivative of the function, which gives us the slope of the tangent at any given point on the curve. The derivative of f(x) is 2x, so at x = 1, the slope (m) of the tangent line is 2(1) = 2. Now that we have the slope, we can use the point-slope equation of a line, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point the line goes through. In this case, the point is (1, f(1)) which is (1, 4). Substituting the values, we get y - 4 = 2(x - 1).
Expanding this equation and writing it in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept, gives us y = 2x + 2. This is the equation of the tangent line at the point (1, f(1)).
To plot the point on the tangent line, pick any value of x other than 1 (to avoid the point of tangency itself). Let's use x = 1.1, and plug it into the equation y = 2(1.1) + 2. The corresponding y-value is 2.2 + 2 = 4.2. Therefore, the point (1.1, 4.2) lies on the tangent line and can be plotted. This line would touch the curve of f(x) at the point (1, 4) and continue in a straight line with a slope of 2.