Graph the equation \( y=x^{2}+12 x+32 \) on the accompanying set of axes. You must plot 5 points including the roots and the vertex.

To graph the equation \( y = x^{2} + 12x + 32 \), let's first identify some key points: roots and vertex. 1. **Finding the vertex**: The vertex of a quadratic function in the form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 12 \), so: \[ x = -\frac{12}{2 \cdot 1} = -6. \] Substitute \( x = -6 \) back into the equation to find \( y \): \[ y = (-6)^{2} + 12(-6) + 32 = 36 - 72 + 32 = -4. \] Thus, the vertex is at the point \((-6, -4)\). 2. **Finding the roots**: To find the roots, we solve the equation \( x^{2} + 12x + 32 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 12 \), and \( c = 32 \): \[ x = \frac{-12 \pm \sqrt{12^{2} - 4 \cdot 1 \cdot 32}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 128}}{2} = \frac{-12 \pm \sqrt{16}}{2} = \frac{-12 \pm 4}{2}. \] This gives two roots: \[ x = \frac{-8}{2} = -4 \quad \text{and} \quad x = \frac{-16}{2} = -8. \] Therefore, the roots are at the points \((-4, 0)\) and \((-8, 0)\). 3. **Choose some additional points**: To complete our five points, we can choose \( x = -7 \) and \( x = -5 \): - For \( x = -7 \): \[ y = (-7)^{2} + 12(-7) + 32 = 49 - 84 + 32 = -3, \] giving the point \((-7, -3)\). - For \( x = -5 \): \[ y = (-5)^{2} + 12(-5) + 32 = 25 - 60 + 32 = -3, \] giving the point \((-5, -3)\). 4. **Summary of points to plot**: - Vertex: \((-6, -4)\) - Roots: \((-4, 0)\) and \((-8, 0)\) - Additional points: \((-7, -3)\) and \((-5, -3)\) Now you can plot these points on the axes and draw a smooth parabola through them, opening upwards! Remember, since the vertex is at \((-6, -4)\), that will be the lowest point of the curve, making a lovely U-shape. Enjoy graphing!