each given function \( f(x) \) : Draw the graphs of \( f(x) \) and its inverse on the same set of axes. Clearly show all the intercepts with the axes, asymptotes and lines of symmetry. Write the equation of the inverse of \( f(x) \) in the form \( y=\ldots \) \( f(x)=\frac{1}{(x+2)}+3 \) \( f(x)=(x-3)^{2}-1 \) \( f(x)=\frac{-2}{(x-4)}+2 \) \( f(x)=-(x+4)^{2}+1 \)

Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=x^{2}-6x+8\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=0^{2}-6\times 0+8\) - step2: Multiply: \(f\left(0\right)=0^{2}-0+8\) - step3: Simplify: \(f\left(0\right)=8\) Analyze the y intercept of the function \( f(x)=-(x+4)^{2}+1 \) Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=-x^{2}-8x-15\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=-0^{2}-8\times 0-15\) - step2: Multiply: \(f\left(0\right)=-0^{2}-0-15\) - step3: Simplify: \(f\left(0\right)=-15\) Analyze the y intercept of the function \( f(x)=\frac{-2}{(x-4)}+2 \) Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=-\frac{2}{x-4}+2\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=-\frac{2}{0-4}+2\) - step2: Simplify: \(f\left(0\right)=\frac{5}{2}\) Analyze the y intercept of the function \( f(x)=\frac{1}{(x+2)}+3 \) Function by following steps: - step0: Find the y-intercept: \(f\left(x\right)=\frac{1}{x+2}+3\) - step1: Set \(x\)=0\(:\) \(f\left(0\right)=\frac{1}{0+2}+3\) - step2: Simplify: \(f\left(0\right)=\frac{7}{2}\) Analyze the x intercept of the function \( f(x)=\frac{1}{(x+2)}+3 \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=\frac{1}{x+2}+3\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=\frac{1}{x+2}+3\) - step2: Swap the sides: \(\frac{1}{x+2}+3=0\) - step3: Find the domain: \(\frac{1}{x+2}+3=0,x\neq -2\) - step4: Move the constant to the right side: \(\frac{1}{x+2}=0-3\) - step5: Remove 0: \(\frac{1}{x+2}=-3\) - step6: Cross multiply: \(1=\left(x+2\right)\left(-3\right)\) - step7: Simplify the equation: \(1=-3\left(x+2\right)\) - step8: Swap the sides: \(-3\left(x+2\right)=1\) - step9: Change the signs: \(3\left(x+2\right)=-1\) - step10: Divide both sides: \(\frac{3\left(x+2\right)}{3}=\frac{-1}{3}\) - step11: Divide the numbers: \(x+2=-\frac{1}{3}\) - step12: Move the constant to the right side: \(x=-\frac{1}{3}-2\) - step13: Subtract the numbers: \(x=-\frac{7}{3}\) - step14: Check if the solution is in the defined range: \(x=-\frac{7}{3},x\neq -2\) - step15: Find the intersection: \(x=-\frac{7}{3}\) Analyze the x intercept of the function \( f(x)=-(x+4)^{2}+1 \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=-x^{2}-8x-15\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=-x^{2}-8x-15\) - step2: Swap the sides: \(-x^{2}-8x-15=0\) - step3: Factor the expression: \(-\left(x+3\right)\left(x+5\right)=0\) - step4: Divide the terms: \(\left(x+3\right)\left(x+5\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x+3=0\\&x+5=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=-3\\&x=-5\end{align}\) - step7: Calculate: \(x_{1}=-5,x_{2}=-3\) Analyze the x intercept of the function \( f(x)=\frac{-2}{(x-4)}+2 \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=-\frac{2}{x-4}+2\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=-\frac{2}{x-4}+2\) - step2: Swap the sides: \(-\frac{2}{x-4}+2=0\) - step3: Find the domain: \(-\frac{2}{x-4}+2=0,x\neq 4\) - step4: Move the constant to the right side: \(-\frac{2}{x-4}=0-2\) - step5: Remove 0: \(-\frac{2}{x-4}=-2\) - step6: Rewrite the expression: \(\frac{-2}{x-4}=-2\) - step7: Cross multiply: \(-2=\left(x-4\right)\left(-2\right)\) - step8: Simplify the equation: \(-2=-2\left(x-4\right)\) - step9: Evaluate: \(1=x-4\) - step10: Swap the sides: \(x-4=1\) - step11: Move the constant to the right side: \(x=1+4\) - step12: Add the numbers: \(x=5\) - step13: Check if the solution is in the defined range: \(x=5,x\neq 4\) - step14: Find the intersection: \(x=5\) Analyze the x intercept of the function \( f(x)=(x-3)^{2}-1 \) Function by following steps: - step0: Find the \(x\)-intercept/zero: \(f\left(x\right)=x^{2}-6x+8\) - step1: Set \(f\left(x\right)\)=0\(:\) \(0=x^{2}-6x+8\) - step2: Swap the sides: \(x^{2}-6x+8=0\) - step3: Factor the expression: \(\left(x-4\right)\left(x-2\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-4=0\\&x-2=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=4\\&x=2\end{align}\) - step6: Calculate: \(x_{1}=2,x_{2}=4\) Solve the equation \( y=-(x+4)^{2}+1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=-x^{2}-8x-15\) - step1: Swap the sides: \(-x^{2}-8x-15=y\) - step2: Move the expression to the left side: \(-x^{2}-8x-15-y=0\) - step3: Multiply both sides: \(x^{2}+8x+15+y=0\) - step4: Solve using the quadratic formula: \(x=\frac{-8\pm \sqrt{8^{2}-4\left(15+y\right)}}{2}\) - step5: Simplify the expression: \(x=\frac{-8\pm \sqrt{4-4y}}{2}\) - step6: Simplify the expression: \(x=\frac{-8\pm 2\sqrt{1-y}}{2}\) - step7: Separate into possible cases: \(\begin{align}&x=\frac{-8+2\sqrt{1-y}}{2}\\&x=\frac{-8-2\sqrt{1-y}}{2}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=-4+\sqrt{1-y}\\&x=\frac{-8-2\sqrt{1-y}}{2}\end{align}\) - step9: Simplify the expression: \(\begin{align}&x=-4+\sqrt{1-y}\\&x=-4-\sqrt{1-y}\end{align}\) Solve the equation \( y=(x-3)^{2}-1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=x^{2}-6x+8\) - step1: Swap the sides: \(x^{2}-6x+8=y\) - step2: Move the expression to the left side: \(x^{2}-6x+8-y=0\) - step3: Solve using the quadratic formula: \(x=\frac{6\pm \sqrt{\left(-6\right)^{2}-4\left(8-y\right)}}{2}\) - step4: Simplify the expression: \(x=\frac{6\pm \sqrt{4+4y}}{2}\) - step5: Simplify the expression: \(x=\frac{6\pm 2\sqrt{1+y}}{2}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{6+2\sqrt{1+y}}{2}\\&x=\frac{6-2\sqrt{1+y}}{2}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=3+\sqrt{1+y}\\&x=\frac{6-2\sqrt{1+y}}{2}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=3+\sqrt{1+y}\\&x=3-\sqrt{1+y}\end{align}\) Solve the equation \( y=\frac{-2}{(x-4)}+2 \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=-\frac{2}{x-4}+2\) - step1: Swap the sides: \(-\frac{2}{x-4}+2=y\) - step2: Move the constant to the right side: \(-\frac{2}{x-4}=y-2\) - step3: Multiply both sides of the equation by LCD: \(-\frac{2}{x-4}\times \left(x-4\right)=\left(y-2\right)\left(x-4\right)\) - step4: Simplify the equation: \(-2=\left(y-2\right)x-4y+8\) - step5: Swap the sides: \(\left(y-2\right)x-4y+8=-2\) - step6: Move the constant to the right side: \(\left(y-2\right)x=-2+4y-8\) - step7: Subtract the numbers: \(\left(y-2\right)x=-10+4y\) - step8: Divide both sides: \(\frac{\left(y-2\right)x}{y-2}=\frac{-10+4y}{y-2}\) - step9: Divide the numbers: \(x=\frac{-10+4y}{y-2}\) Solve the equation \( y=\frac{1}{(x+2)}+3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(y=\frac{1}{x+2}+3\) - step1: Swap the sides: \(\frac{1}{x+2}+3=y\) - step2: Move the constant to the right side: \(\frac{1}{x+2}=y-3\) - step3: Multiply both sides of the equation by LCD: \(\frac{1}{x+2}\times \left(x+2\right)=\left(y-3\right)\left(x+2\right)\) - step4: Simplify the equation: \(1=\left(y-3\right)x+2y-6\) - step5: Swap the sides: \(\left(y-3\right)x+2y-6=1\) - step6: Move the constant to the right side: \(\left(y-3\right)x=1-\left(2y-6\right)\) - step7: Subtract the terms: \(\left(y-3\right)x=7-2y\) - step8: Divide both sides: \(\frac{\left(y-3\right)x}{y-3}=\frac{7-2y}{y-3}\) - step9: Divide the numbers: \(x=\frac{7-2y}{y-3}\) Let's analyze each function \( f(x) \) step by step, including finding the intercepts, asymptotes, and the equations of their inverses. ### 1. Function: \( f(x) = \frac{1}{(x+2)} + 3 \) **Intercepts:** - **X-intercept:** Set \( f(x) = 0 \): \[ 0 = \frac{1}{(x+2)} + 3 \implies x = -\frac{7}{3} \] - **Y-intercept:** Set \( x = 0 \): \[ f(0) = \frac{1}{(0+2)} + 3 = \frac{1}{2} + 3 = \frac{7}{2} \text{ or } 3.5 \] **Asymptotes:** - Vertical asymptote at \( x = -2 \) (where the denominator is zero). **Inverse:** To find the inverse, swap \( x \) and \( y \): \[ y = \frac{1}{(x+2)} + 3 \implies x = \frac{1}{(y+2)} + 3 \implies y = \frac{7 - 2x}{x - 3} \] ### 2. Function: \( f(x) = (x-3)^{2} - 1 \) **Intercepts:** - **X-intercepts:** Set \( f(x) = 0 \): \[ 0 = (x-3)^{2} - 1 \implies x = 2, 4 \] - **Y-intercept:** Set \( x = 0 \): \[ f(0) = (0-3)^{2} - 1 = 9 - 1 = 8 \] **Asymptotes:** - No vertical or horizontal asymptotes. **Inverse:** To find the inverse, swap \( x \) and \( y \): \[ y = (x-3)^{2} - 1 \implies x = (y-3)^{2} - 1 \implies y = 3 + \sqrt{1+y} \text{ or } y = 3 - \sqrt{1+y} \] ### 3. Function: \( f(x) = \frac{-2}{(x-4)} + 2 \) **Intercepts:** - **X-intercept:** Set \( f(x) = 0 \): \[ 0 = \frac{-2}{(x-4)} + 2 \implies x = 5 \] - **Y-intercept:** Set \( x = 0 \): \[ f(0) = \frac{-2}{(0-4)} + 2 = \frac{1}{2} + 2 = \frac{5}{2} \text{ or } 2.5 \] **Asymptotes:** - Vertical asymptote at \( x = 4 \). **Inverse:** To find the inverse, swap \( x \) and \( y \): \[ y = \frac{-2}{(x-4)} + 2 \implies x = \frac{-2}{(y-4)} + 2 \implies y = \frac{-10 + 4x}{x - 2} \] ### 4. Function: \( f(x) = -(x+4)^{2} + 1 \) **Intercepts:** - **X-intercepts:** Set \( f(x) = 0 \): \[ 0 = -(x+4)^{2} + 1 \implies x = -5, -3 \] - **Y-intercept:** Set \( x = 0 \): \[ f(0) = -(0+4)^{2} + 1 = -16 + 1 = -15 \] **Asymptotes:** - No vertical or horizontal asymptotes. **Inverse:** To find the inverse, swap \( x \) and \( y \): \[ y = -(x+4)^{2} + 1 \implies x = -(y+4)^{2} + 1 \implies y = -4 + \sqrt{1-y} \text{ or } y = -4 - \sqrt{1-y} \] ### Summary of Inverses 1. \( f^{-1}(x) = \frac{7 - 2x}{x - 3} \) 2. \( f^{-1}(x) = 3 \pm \sqrt{1+x} \) 3. \( f^{-1}(x) = \frac{-10 + 4x}{x - 2} \) 4. \( f^{-1}(x) = -4 \pm \sqrt{1+x} \) ### Graphing To graph these functions and their inverses, plot the functions along with their intercepts and asymptotes. The line \( y = x \) will serve as the line of symmetry for the function and its inverse. If you need the actual graphs plotted, please let me know!