QUESTION 2 \( f(x)=-2 x^{2}+2 \) and \( g(x)=2^{x}+1 \) are the defining equations of graphs \( f \) and \( g \) respectively. Write down an equation for the asymptote of \( g \). 2.1 Calculate the coordinates of the turning point of \( f \). 2.3 Sketch the graphs of \( f \) and \( g \) on the same set of axes, clearly showing ALL intercepts with the axes, turning points and asymptotes. 2.4 Write down the range of \( f \). 2.5 What transformation does the graph of \( y=f(x) \) undergo in order to obtain the graph of \( y=2 x^{2}-2 \) ?

Let's go through each part of the question step by step. ### 2.1 Equation for the asymptote of \( g \) The function \( g(x) = 2^x + 1 \) is an exponential function. As \( x \) approaches negative infinity, \( 2^x \) approaches 0. Therefore, the horizontal asymptote of \( g(x) \) is: \[ y = 1 \] ### 2.2 Calculate the coordinates of the turning point of \( f \) The function \( f(x) = -2x^2 + 2 \) is a quadratic function. The turning point (vertex) of a quadratic function in the form \( ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] In this case, \( a = -2 \) and \( b = 0 \): \[ x = -\frac{0}{2 \cdot -2} = 0 \] Now, substitute \( x = 0 \) back into \( f(x) \) to find the y-coordinate of the turning point: \[ f(0) = -2(0)^2 + 2 = 2 \] Thus, the coordinates of the turning point of \( f \) are: \[ (0, 2) \] ### 2.3 Sketch the graphs of \( f \) and \( g \) To sketch the graphs, we need to find the intercepts and the turning point for \( f \) and the asymptote for \( g \). **For \( f(x) = -2x^2 + 2 \):** - **Y-intercept:** Set \( x = 0 \): \[ f(0) = 2 \quad \text{(Y-intercept: (0, 2))} \] - **X-intercepts:** Set \( f(x) = 0 \): \[ -2x^2 + 2 = 0 \implies 2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1 \] (X-intercepts: (1, 0) and (-1, 0)) - **Turning point:** (0, 2) **For \( g(x) = 2^x + 1 \):** - **Y-intercept:** Set \( x = 0 \): \[ g(0) = 2^0 + 1 = 2 \quad \text{(Y-intercept: (0, 2))} \] - **X-intercept:** Set \( g(x) = 0 \): \[ 2^x + 1 = 0 \quad \text{(no real solution)} \] - **Asymptote:** \( y = 1 \) **Sketch:** - The graph of \( f \) is a downward-opening parabola with a vertex at (0, 2) and intercepts at (1, 0) and (-1, 0). - The graph of \( g \) approaches the horizontal asymptote \( y = 1 \) as \( x \) approaches negative infinity and passes through (0, 2). ### 2.4 Write down the range of \( f \) The function \( f(x) = -2x^2 + 2 \) opens downwards, and its maximum value (the y-coordinate of the turning point) is 2. As \( x \) moves away from 0, \( f(x) \) decreases without bound. Therefore, the range of \( f \) is: \[ (-\infty, 2] \] ### 2.5 What transformation does the graph of \( y=f(x) \) undergo to obtain the graph of \( y=2x^2-2 \)? The function \( y = 2x^2 - 2 \) can be obtained from \( y = f(x) = -2x^2 + 2 \) through the following transformations: 1. **Reflection:** Reflect \( f(x) \) across the x-axis (changing the sign of the coefficient of \( x^2 \)). 2. **Vertical Stretch:** Multiply by -1 to get \( 2x^2 \) (which is a vertical stretch by a factor of 1). 3. **Vertical Shift:** Shift down by 4 units (from +2 to -2). In summary, the transformation involves reflecting across the x-axis, stretching vertically, and shifting downwards.