Let's fill out the chart using the two quadratic equations: ### 1. Complete the Chart | Equation | \( y = x^2 - x - 30 \) | \( y = -5(x - 1)(x + 3) \) | |-------------------------|-----------------------------|--------------------------------------| | **Direction of Opening**| Upwards (a = +1) | Downwards (a = -5) | | **Maximum or Minimum** | Minimum | Maximum | | **Vertex** | \( x = \frac{-b}{2a} = \frac{1}{2} \); \( y = (\frac{1}{2})^2 - (\frac{1}{2}) - 30 = -30.25 \)<br>Vertex: \( (\frac{1}{2}, -30.25) \) | x-coordinate: \( x = \frac{1 + (-3)}{2} = -1 \)<br>y-coordinate: \( y = -5(-1-1)(-1+3) = -5(-2)(2) = 20 \)<br>Vertex: \( (-1, 20) \) | | **Equation of Axis of Symmetry** | \( x = \frac{1}{2} \) | \( x = -1 \) | | **x-intercepts** | Factor: \( (x - 6)(x + 5) = 0 \)<br>Intercepts: \( x = 6, -5 \) | Set each factor to 0:<br>\( x-1=0 \Rightarrow x=1 \)<br>\( x+3=0 \Rightarrow x=-3 \)<br>Intercepts: \( x = 1, -3 \) | | **y-intercept** | Set \( x = 0 \): \( y = -30 \) | \( x = 0 \Rightarrow y = -5(-1)(3) = 15 \) | | **Domain** | \( (-\infty, \infty) \) | \( (-\infty, \infty) \) | | **Range** | \( [-30.25, \infty) \) | \( (-\infty, 20] \) | --- ### 2. Explanation (vertex and max/min) The vertex of the quadratic is the highest or lowest point (maximum or minimum) on the graph, depending on the direction the parabola opens. For \( y = ax^2 + bx + c \): - If \( a > 0 \), the parabola opens upwards and the vertex is the minimum. - If \( a < 0 \), it opens downwards and the vertex is the maximum. The x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Substitute this x-value into the equation to get the y-coordinate (max or min value). --- Would you like help with the rest of the questions?

Certainly! Let's go through each part step-by-step. --- ### **Question 1: Complete the chart using the two equations** Given equations: - \( y = x^2 - x - 30 \) - \( y = -5(x - 1)(x + 3) \) --- ### **Part 1a: Find the vertex of \( y = x^2 - x - 30 \)** **Step 1:** Write the quadratic in standard form: Already given as \( y = x^2 - x - 30 \). **Step 2:** Find the x-coordinate of the vertex using the formula: \[ x = -\frac{b}{2a} \] where \( a = 1 \), \( b = -1 \). \[ x = -\frac{-1}{2 \times 1} = \frac{1}{2} \] **Step 3:** Find the y-coordinate by substituting \( x = \frac{1}{2} \) into the equation: \[ y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 30 \] Calculate step-by-step: \[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ y = \frac{1}{4} - \frac{1}{2} - 30 \] Convert all to quarters: \[ \frac{1}{4} - \frac{2}{4} - 30 = -\frac{1}{4} - 30 \] Express 30 as quarters: \[ - \frac{1}{4} - \frac{120}{4} = -\frac{121}{4} = -30.25 \] **Vertex:** \( \left( \frac{1}{2}, -30.25 \right) \) **Step 4:** Equation of the axis of symmetry: \[ x = \frac{1}{2} \] --- ### **Part 1b: Find the vertex of \( y = -5(x - 1)(x + 3) \)** **Step 1:** Expand the equation: \[ y = -5 (x - 1)(x + 3) \] First, expand the factors: \[ (x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3 \] Multiply by -5: \[ y = -5(x^2 + 2x - 3) = -5x^2 - 10x + 15 \] **Step 2:** Find the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] Here, \( a = -5 \), \( b = -10 \): \[ x = -\frac{-10}{2 \times -5} = \frac{10}{-10} = -1 \] **Step 3:** Find y-coordinate by substituting \( x = -1 \): \[ y = -5(-1)^2 - 10(-1) + 15 \] Calculate step-by-step: \[ -5(1) = -5 \] \[ -10(-1) = +10 \] So, \[ y = -5 + 10 + 15 = 20 \] **Vertex:** \( (-1, 20) \) **Step 4:** Equation of the axis of symmetry: \[ x = -1 \] --- ### **Summary for Question 1:** | Equation | Direction | Vertex | Axis of Symmetry | x-intercepts | y-intercept | Domain | Range | |------------|--------------|---------|------------------|--------------|--------------|---------|--------| | \( y = x^2 - x - 30 \) | Upward | \( \left( \frac{1}{2}, -30.25 \right) \) | \( x = \frac{1}{2} \) | \( x = 6, -5 \) | \( y = -30 \) | \( (-\infty, \infty) \) | \( [-30.25, \infty) \) | | \( y = -5(x - 1)(x + 3) \) | Downward | \( (-1, 20) \) | \( x = -1 \) | \( x = 1, -3 \) | \( y = 15 \) | \( (-\infty, \infty) \) | \( (-\infty, 20] \) | --- ### **Question 2: How do the coordinates of the vertex help find the maximum or minimum value?** **Explanation:** - The **vertex** of a parabola represents its highest or lowest point. - If the parabola opens **upward** (\( a > 0 \)), the vertex is the **minimum** point. - If it opens **downward** (\( a < 0 \)), the vertex is the **maximum** point. - The **x-coordinate** of the vertex is found using \( x = -\frac{b}{2a} \). Plugging this x-value into the quadratic gives the **y-value** (this is the maximum or minimum y). - This y-value is the **maximum or minimum value** of the quadratic function. --- ### **Question 3: Sketch the graph of \( y = x^2 - x - 30 \)** **Steps to sketch:** - Plot the vertex at \( \left( \frac{1}{2}, -30.25 \right) \). - Find and plot the **x-intercepts** at \( x = 6 \) and \( x = -5 \). - Plot the **y-intercept** at \( y = -30 \) when \( x = 0 \). - Since \( a = 1 > 0 \), the parabola opens upward. - Draw a symmetric U-shaped curve passing through these points. --- ### **Question 4: Find the vertex of \( y = (5 - x)(x - 15) \)** **Step 1:** Expand the factored form: \[ y = (5 - x)(x - 15) \] \[ = 5x - 75 - x^2 + 15x \] \[ = -x^2 + (5x + 15x) - 75 = -x^2 + 20x - 75 \] **Step 2:** Find the x-coordinate of the vertex: \[ x = -\frac{b}{2a} \] where \( a = -1 \), \( b = 20 \): \[ x = -\frac{20}{2 \times -1} = -\frac{20}{-2} = 10 \] **Step 3:** Find y-coordinate: \[ y = - (10)^2 + 20 \times 10 - 75 = -100 + 200 - 75 = 25 \] **Vertex:** \( (10, 25) \) --- ### **Question 5: Rocket height problem** **Given:** - Rocket is fired off the ground (height = 0 at \( x = 0 \)) - Reaches a height of 250 meters at some point - Target is 200 meters away (horizontal distance) **Modeling the path:** A quadratic function models the parabolic trajectory: \[ h(t) = -a t^2 + v t + h_0 \] Where: - \( h(t) \) = height at time \( t \) - \( h_0 = 0 \) (ground level) - \( v \) = initial velocity component (unknown) - \( a \) = gravity effect (positive constant) Since the maximum height is 250 m, and it is reached at the vertex, the vertex height is 250. **Assuming:** - The parabola opens downward. - The maximum height occurs at \( t = t_{max} \). Without specific initial velocity and time data, a more detailed model isn't possible here, but generally: - The maximum height (vertex y-value) is 250 meters. - The horizontal distance (range) is 200 meters. This indicates the parabola peaks at 250 meters, roughly halfway to the target (assuming symmetrical flight), which can be used to estimate initial velocity and time to reach maximum height. --- **Would you like me to help set up the actual quadratic equation for the rocket's path?** --- **Let me know if you'd like further clarification or assistance!**