QUESTION 3 The equation of a hyperbola is given by \( f(x)=\frac{3}{x-7}-4 \). Write down the equation of the new function that is formed when \( f \) 3.1 Shift two units to the left 3.2 Shift 3 units up Shift 1 unit right and 2 units down 3.4 The equation of the new hyperbola has new asymptotes at \( x \) QUESTION 4 Sketch on the same set of axes the graphs of \( f(x)=-2 x^{2}-4 x+6 \) and Clearly indicate all intercepts with the axes, turning point(s) and asymp
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When you shift the function \( f(x) = \frac{3}{x-7} - 4 \) two units to the left, you essentially replace \( x \) with \( x + 2 \). This gives you the new function \( f(x) = \frac{3}{(x + 2) - 7} - 4 = \frac{3}{x - 5} - 4 \). For the shift of 3 units up, simply add 3 to the entire function, resulting in \( f(x) = \frac{3}{x - 7} - 1 \). When you shift the function 1 unit right and 2 units down, replace \( x \) with \( x - 1 \) and modify the constant: \( f(x) = \frac{3}{(x - 1) - 7} - 6 = \frac{3}{x - 8} - 6 \). The new asymptotes for this hyperbola move with the shifts; the vertical asymptote is now at \( x = 8 \). To sketch the graph of the parabola \( f(x) = -2x^2 - 4x + 6 \), first find the x-intercepts using the quadratic formula, which yields points at \( x = -1 \) and \( x = 3 \). The turning point, or vertex, can be found at \( x = -b/(2a) = 1 \), which gives \( f(1) = 4 \), making the turning point at \( (1, 4) \). The y-intercept is at \( (0, 6) \) with the curve opening downwards, so be sure to highlight these points when sketching!
