The graph of a transformed exponential function has the following characteristics:
horizontal asymptote at y = -9 passes through the points (-4, -8) and (-1, 18)
What are the coordinates of the x-intercept?
Formula:
Transformed Exponential Function in the Form \(y = b^{x–h} + k\)
h = Horizontal Shift
k = Vertical Shift
Horizontal asymptote at \(y = -9, \text{ so } k = -9\)
\(\begin{array}{|lrcll|} \hline \text{Point 1 }~ (-4,-8): & y &=& b^{x–h} - 9 \\ & -8 &=& b^{-4–h} - 9 \\ & 1 &=& b^{-4–h} \quad & | \quad \ln() \text{ both sides}\\ & \ln(1) &=& \ln(b^{-4–h}) \\ & \ln(1) &=& (-4–h)\cdot \ln(b) \quad & | \quad \ln(1) = 0 \\ & 0 &=& (-4–h)\cdot \ln(b) \quad & | \quad : \ln(b) \\ & 0 &=& -4–h \quad & | \quad +h \\ & \mathbf{h} & \mathbf{=} & \mathbf{ -4 } \\\\ \text{Point 2 }~ (-1,18): & y &=& b^{x–h} - 9 \\ & 18 &=& b^{-1–h} - 9 \\ & 27 &=& b^{-1–h} \quad & | \quad h = -4 \\ & 27 &=& b^{-1+4} \\ & 27 &=& b^{3} \quad & | \quad 27 = 3^3 \\ & 3^3 &=& b^{3} \quad & | \quad \sqrt[3]{} \text{ both sides}\\ & 3 &=& b \\ & \mathbf{b} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)
The Transformed Exponential Function is \(y = 3^{x+4} - 9\)
The coordinates of the x-intercept:
\(\begin{array}{|rcll|} \hline y &=& 3^{x+4} - 9 \quad & | \quad \mathbf{y = 0} \\ 0 &=& 3^{x+4} - 9 \quad & | \quad +9 \\ 9 &=& 3^{x+4} \quad & | \quad 9 = 3^2 \\ 3^2 &=& 3^{x+4} \\ 2 &=& x+4 \quad & | \quad -4 \\ 2-4 &=& x \\ -2 &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{-2} \\ \hline \end{array}\)
The coordinates of the x-intercept is (-2,0)
The graph is:
