Question Plot the given parabola on the axes. Plot the roots, the vertex and two other points. \[ y=x^{2}+20 x+84 \]

The given parabola can be expressed in the standard form \(y = ax^2 + bx + c\) where \(a = 1\), \(b = 20\), and \(c = 84\). To find the roots, you can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculating the discriminant: \(b^2 - 4ac = 20^2 - 4(1)(84) = 400 - 336 = 64\). This positive value indicates two real roots. Now, finding the roots: 1. \(x_1 = \frac{-20 + 8}{2} = -6\) 2. \(x_2 = \frac{-20 - 8}{2} = -14\) Next, to find the vertex, use the formula \(x_v = -\frac{b}{2a} = -\frac{20}{2(1)} = -10\). We can then find the corresponding y-coordinate: \(y_v = (-10)^2 + 20(-10) + 84 = 100 - 200 + 84 = -16\), making the vertex \((-10, -16)\). Now choose two additional points such as: - For \(x= -12\): \(y = (-12)^2 + 20(-12) + 84 = 144 - 240 + 84 = -12\). - For \(x= -8\): \(y = (-8)^2 + 20(-8) + 84 = 64 - 160 + 84 = -12\). So you can plot these key points: - Roots: \((-14, 0)\) and \((-6, 0)\). - Vertex: \((-10, -16)\). - Extra points: \((-12, -12)\) and \((-8, -12)\). Now, draw the parabola opening upwards and connect the dots with a smooth curve!