QUESTION 4 Sketch on the same set of axes the graphs of \( f(x)=-2 x^{2}-4 x+6 \) and \( g(x)=-2 \cdot 2^{x-1}+1 \) Clearly indicate all intercepts with the axes, turning point(s) and asymptote(s).
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Mind Expander
To create an engaging and informative response, let's take a look at the characteristics of both functions, touching on intercepts and asymptotes! First, the function \( f(x) = -2x^2 - 4x + 6 \) is a downward-opening parabola. Its y-intercept can be found by substituting \( x = 0 \), yielding \( f(0) = 6 \). The x-intercepts can be found through factoring or the quadratic formula, revealing intercepts at points \( (-3, 0) \) and \( (0, 0) \). The vertex, or turning point, can be located using the formula \( x = -\frac{b}{2a} \), which gives \( x = -1 \). Substituting \( x = -1 \) back into the function reveals the vertex at \( (-1, 8) \). Now, on to \( g(x) = -2 \cdot 2^{x-1} + 1 \)! This function is an exponential curve. The y-intercept can be found by evaluating \( g(0) = -2 \cdot 2^{-1} + 1 = 0 \), so it crosses at the point \( (0, 0) \). As \( x \to -\infty \), \( g(x) \) approaches the horizontal asymptote \( y = 1 \). This means that although \( g(x) \) climbs up after hitting its intercept, it will never actually reach the line \( y = 1 \). You've got a colorful intersection of a quadratic and an exponential function on the same graph!