The area of the triangle formed by the \( x \)-and \( y \)-intercepts of the parabola \( y=0.5(x-3)(x+k) \) is equal to 1.5 square units. Find all possible values of \( k \)

To find the area of the triangle formed by the \( x \)-and \( y \)-intercepts of the parabola \( y=0.5(x-3)(x+k) \), we first need to find the \( x \)-intercepts by setting \( y \) to 0 and solving for \( x \). Then, we can find the \( y \)-intercept by setting \( x \) to 0 and solving for \( y \). Finally, we can calculate the area of the triangle formed by these intercepts. 1. Find the \( x \)-intercepts: Setting \( y = 0 \) in the equation \( y = 0.5(x-3)(x+k) \): \[ 0 = 0.5(x-3)(x+k) \] 2. Find the \( y \)-intercept: Setting \( x = 0 \) in the equation \( y = 0.5(x-3)(x+k) \): \[ y = 0.5(0-3)(0+k) \] 3. Calculate the area of the triangle formed by the intercepts: The area of a triangle formed by two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Let's calculate the \( x \)-intercepts first. Solve the equation by following steps: - step0: Solve for \(x\): \(0=0.5\left(x-3\right)\left(x+k\right)\) - step1: Swap the sides: \(0.5\left(x-3\right)\left(x+k\right)=0\) - step2: Calculate: \(\left(x-3\right)\left(x+k\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-3=0\\&x+k=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=3\\&x=-k\end{align}\) It seems there was an error in the substitution. Let's correct it and find the \( y \)-intercept by setting \( x = 0 \) in the equation \( y = 0.5(x-3)(x+k) \). Evaluate the expression by following steps: - step0: Evaluate: \(0.5\left(x-3\right)\left(x+k\right)\) - step1: Substitute: \(0.5\left(0-3\right)\left(0+k\right)\) - step2: Remove 0: \(0.5\left(-3\right)\left(0+k\right)\) - step3: Remove 0: \(0.5\left(-3\right)k\) - step4: Rewrite the expression: \(-0.5\times 3k\) - step5: Multiply the terms: \(-1.5k\) The \( y \)-intercept of the parabola is \( y = -1.5k \). Now, we have the \( x \)-intercepts as \( x = 3 \) and \( x = -k \), and the \( y \)-intercept as \( y = -1.5k \). The area of the triangle formed by these intercepts is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] The base of the triangle is the distance between the \( x \)-intercepts, which is \( |3 - (-k)| = |3 + k| \), and the height is the absolute value of the \( y \)-intercept, which is \( |-1.5k| = 1.5|k| \). Given that the area of the triangle is 1.5 square units, we have: \[ 1.5 = \frac{1}{2} \times |3 + k| \times 1.5|k| \] Now, we can solve this equation to find all possible values of \( k \). Solve the equation by following steps: - step0: Solve for \(k\): \(1.5=\frac{1}{2}\left|3+k\right|\times 1.5\left|k\right|\) - step1: Multiply the terms: \(1.5=\frac{3}{4}\left|k\left(3+k\right)\right|\) - step2: Swap the sides: \(\frac{3}{4}\left|k\left(3+k\right)\right|=1.5\) - step3: Multiply by the reciprocal: \(\frac{3}{4}\left|k\left(3+k\right)\right|\times \frac{4}{3}=1.5\times \frac{4}{3}\) - step4: Multiply: \(\left|k\left(3+k\right)\right|=2\) - step5: Separate into possible cases: \(\begin{align}&k\left(3+k\right)=2\\&k\left(3+k\right)=-2\end{align}\) - step6: Solve the equation: \(\begin{align}&k=\frac{-3+\sqrt{17}}{2}\\&k=-\frac{3+\sqrt{17}}{2}\\&k=-1\\&k=-2\end{align}\) - step7: Rewrite: \(k_{1}=-\frac{3+\sqrt{17}}{2},k_{2}=-2,k_{3}=-1,k_{4}=\frac{-3+\sqrt{17}}{2}\) The possible values of \( k \) that satisfy the given condition are: \[ k = -\frac{3+\sqrt{17}}{2}, -2, -1, \frac{-3+\sqrt{17}}{2} \]