Questions: Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation for the parabola's axis of the function's domain and range. f(x)=x^2+16x+8 (-8,-56) (Type an ordered pair.) What are the x-intercepts? (-8+2sqrt(14), 0),(-8-2sqrt(14), 0) What is the y-intercept? (Type an ordered pair.)

Solution Steps

The vertex form of a quadratic function \( f(x) = ax^2 + bx + c \) is given by: \[ x = -\frac{b}{2a} \] For the given function \( f(x) = x^2 + 16x + 8 \), we have \( a = 1 \) and \( b = 16 \). Therefore, \[ x = -\frac{16}{2 \cdot 1} = -8 \] To find the y-coordinate of the vertex, substitute \( x = -8 \) into the function: \[ f(-8) = (-8)^2 + 16(-8) + 8 = 64 - 128 + 8 = -56 \] Thus, the vertex is \((-8, -56)\).

The x-intercepts are found by solving \( f(x) = 0 \): \[ x^2 + 16x + 8 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{-16 \pm \sqrt{256 - 32}}{2} = \frac{-16 \pm \sqrt{224}}{2} = \frac{-16 \pm 4\sqrt{14}}{2} = -8 \pm 2\sqrt{14} \] Thus, the x-intercepts are \((-8 + 2\sqrt{14}, 0)\) and \((-8 - 2\sqrt{14}, 0)\).

The y-intercept is found by evaluating \( f(x) \) at \( x = 0 \): \[ f(0) = 0^2 + 16 \cdot 0 + 8 = 8 \] Thus, the y-intercept is \((0, 8)\).

Final Answer