QUESTION 2 FUNCTIONS 2.1 Give: \( y=a .2^{x+3}-4 \) 2.1.1 Determine \( a \) if \( (-2,-2) \) is a point on graph. 2.1.2 Write down the equation of the asymptote. (1) 2.1.3 Determine the \( x \) - and \( y \)-intercepts. (4) 2.1.4 Draw a neat graph of \( y=a .2^{x+3}-4 \)

To determine \( a \) given that the point \((-2, -2)\) lies on the graph of the equation \( y = a \cdot 2^{x+3} - 4 \), we can substitute \(-2\) for \(x\) and \(-2\) for \(y\) in the equation. ### 2.1.1 Determine \( a \) Substituting in the point: \[ -2 = a \cdot 2^{-2+3} - 4 \] This simplifies to: \[ -2 = a \cdot 2^{1} - 4 \] \[ -2 = 2a - 4 \] \[ 2a = 2 \implies a = 1 \] ### 2.1.2 Write down the equation of the asymptote. The equation of the horizontal asymptote for this function, as the exponent approaches negative infinity, is given by the vertical shift in the equation. Here, since we subtract 4, the asymptote is: \[ y = -4 \] ### 2.1.3 Determine the \( x \) - and \( y \)-intercepts. - **Finding the \( y \)-intercept**: Set \( x = 0 \). \[ y = 1 \cdot 2^{0 + 3} - 4 = 1 \cdot 2^{3} - 4 = 8 - 4 = 4 \] So, the \( y \)-intercept is \( (0, 4) \). - **Finding the \( x \)-intercept**: Set \( y = 0 \). \[ 0 = 1 \cdot 2^{x + 3} - 4 \implies 1 \cdot 2^{x + 3} = 4 \] \[ 2^{x + 3} = 4 \implies 2^{x + 3} = 2^{2} \] This implies: \[ x + 3 = 2 \implies x = -1 \] So, the \( x \)-intercept is \( (-1, 0) \). ### 2.1.4 Draw a neat graph of \( y = 2^{x + 3} - 4 \) To graph this function, you’ll start with the horizontal asymptote at \( y = -4 \). Plot the intercepts at \( (0, 4) \) and \( (-1, 0) \). Then, observe that as \( x \to -\infty \), \( y \) approaches \( -4 \), and as \( x \to \infty \), \( y \) increases towards positive infinity. Connect these points smoothly to create the graph, representing the exponential function. The graph should show a rapid increase after the x-intercept and flattening out near the line \( y = -4 \).