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Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necessary). \[ y=x^{2}+10 x-96 \]

Ask by Ross Joseph. in the United States
Feb 03,2025

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Answer

The roots of the equation \( y = x^2 + 10x - 96 \) are \( x = 6.000 \) and \( x = -16.000 \). The vertex of the quadratic is at \( (-5.000, -121.000) \).

Solution

To find the roots and vertex of the quadratic function \( y = x^2 + 10x - 96 \), we'll use algebraic methods. ### **1. Finding the Roots** The roots of the quadratic equation \( y = 0 \) can be found using the **quadratic formula**: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the given equation \( x^2 + 10x - 96 = 0 \): - **\( a = 1 \)** - **\( b = 10 \)** - **\( c = -96 \)** **Plugging the values into the quadratic formula:** \[ x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-96)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 384}}{2} = \frac{-10 \pm \sqrt{484}}{2} \] \[ \sqrt{484} = 22 \] So, \[ x = \frac{-10 \pm 22}{2} \] This gives two solutions: 1. \( x = \frac{-10 + 22}{2} = \frac{12}{2} = 6.000 \) 2. \( x = \frac{-10 - 22}{2} = \frac{-32}{2} = -16.000 \) ### **2. Finding the Vertex** The vertex \((h, k)\) of a quadratic function \( y = ax^2 + bx + c \) can be found using: \[ h = -\frac{b}{2a} \] \[ k = y(h) = a(h)^2 + b(h) + c \] For our equation: \[ h = -\frac{10}{2(1)} = -5.000 \] Now, plug \( h = -5.000 \) into the original equation to find \( k \): \[ k = (-5)^2 + 10(-5) - 96 = 25 - 50 - 96 = -121.000 \] ### **Summary of Results** - **Roots:** - \( x = 6.000 \) - \( x = -16.000 \) - **Vertex:** - \( (-5.000, -121.000) \) --- **Final Answer:** - **Roots:** \( x =\ 6.000 \) and \( x =\ -16.000 \) - **Vertex:** \( (-5.000,\ -121.000) \)

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Bonus Knowledge

To find the roots of the quadratic equation \( y = x^{2} + 10x - 96 \), you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, \( a = 1 \), \( b = 10 \), and \( c = -96 \). Plugging these values into the formula, you get: \[ x = \frac{-10 \pm \sqrt{(10)^2 - 4(1)(-96)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 384}}{2} = \frac{-10 \pm \sqrt{484}}{2} = \frac{-10 \pm 22}{2} \] Calculating both roots: \[ x_1 = \frac{12}{2} = 6 \quad \text{and} \quad x_2 = \frac{-32}{2} = -16 \] So the roots are \( x = 6 \) and \( x = -16 \). Now, to find the vertex, use the formula \( x = -\frac{b}{2a} \): \[ x = -\frac{10}{2 \cdot 1} = -5 \] Now substitute \( x = -5 \) back into the equation to find the \( y \)-coordinate: \[ y = (-5)^{2} + 10(-5) - 96 = 25 - 50 - 96 = -121 \] Thus, the vertex is \( (-5, -121) \). Final results: - Roots: \( (6, 0) \) and \( (-16, 0) \) - Vertex: \( (-5, -121) \)

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