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 $f(x) = 2^{x} + 1$ can be sketched as follows:

• Intercepts with axes:
  • Y-intercept: (0; 2)
  • X-intercept: none (since $2^{x} + 1$ is always greater than 1).
• Asymptotes:
  • There is a horizontal asymptote at $y = 1$, as $x$ approaches negative infinity.

The graph should demonstrate exponential growth as $x$ increases, starting near the asymptote at $y=1$.

The average gradient between the points can be calculated using:

$ext{Average Gradient} = \frac{f(1) - f(-2)}{1 - (-2)}$

Calculating the function values:

• $f(1) = 2^{1} + 1 = 3$
• $f(-2) = 2^{-2} + 1 = \frac{1}{4} + 1 = \frac{5}{4}$

Now substituting these values:

$ext{Average Gradient} = \frac{3 - \frac{5}{4}}{1 + 2} = \frac{12/4 - 5/4}{3} = \frac{7/4}{3} = \frac{7}{12}$

Since the asymptote of $f$ is $y = 1$, when we multiply by 3 for $h(x)$, the asymptote shifts accordingly. Therefore, the asymptote of $h(x)=3f(x)$ will be:

$y = 3 \cdot 1 = 3$

Thus, the equation of the asymptote of $h$ is $y = 3$.