Given: f(x) = 2^x + 1 4.1 Determine the coordinates of the y-intercept of f - NSC Mathematics - Question 4 - 2016 - Paper 1

The graph of the function f(x) = 2^x + 1 has the following characteristics:

• Y-Intercept: (0; 2)
• X-Intercept: Since f(x) = 0 has no solution for the equation 2^x + 1 = 0, there is no x-intercept.
• Asymptote: The horizontal asymptote occurs at y = 1, as the function approaches y = 1 as x approaches negative infinity.

The graph is an exponential function that increases without bound as x increases, and it approaches y = 1 but never touches it.

To find the average gradient between x = -2 and x = 1, we need to calculate the values of f at these points:

$f(-2) = 2^{-2} + 1 = \frac{1}{4} + 1 = \frac{5}{4}$

$f(1) = 2^1 + 1 = 2 + 1 = 3$

Now, we can use the formula for the average gradient:

$\text{Average Gradient} = \frac{f(1) - f(-2)}{1 - (-2)} = \frac{3 - \frac{5}{4}}{3} = \frac{\frac{12}{4} - \frac{5}{4}}{3} = \frac{\frac{7}{4}}{3} = \frac{7}{12}$

Since the asymptote of f is y = 1, the asymptote of h(x) = 3f(x) will be:

y = 3(1) = 3.