Final answer:
To find the slope of the tangent line to the graph at x=3, differentiate the function f(x)=x² + 5 to get f'(x)=2x, then evaluate it at x=3 to get a slope of 6. The equation of the tangent line is y=6x-4 using the point-slope form with the point (3, f(3))=(3,14).
Step-by-step explanation:
To find the slope of the tangent line to the graph of the function f(x) = x² + 5 at x = 3, we first need to calculate the derivative of the function. The derivative gives us the slope of the tangent line at any point on the curve. Differentiating f(x) with respect to x, we get f'(x) = 2x. Then, evaluating the derivative at x = 3, we obtain the slope of the tangent line, which is f'(3) = 2(3) = 6.
The equation of the tangent line can be found using the point-slope form, which is y - y1 = m(x - x1) where m is the slope and (x1, y1) is the point of tangency. Since the curve passes through the point (3, f(3)), which is (3, 14), and we have found the slope to be 6, the equation of the tangent line is y - 14 = 6(x - 3). Simplifying this, we get y = 6x - 4, which is the equation of the tangent line at x = 3.