Question: For the quadratic function \( f(x) = 4x^2 + 4x - 4 \), find the vertex, the axis of symmetry, and…

For the quadratic function \( f(x) = 4x^2 + 4x - 4 \), find the vertex, the axis of symmetry, and the \( x \)-intercept(s). Express the vertex as a coordinate pair. If there is more than one intercept, enter as a comma-separated list of points.

• vertex: \((-0.5, -5)\)
• axis of symmetry: \( x = -0.5 \)
• \( x \)-intercept(s):

Solution

To solve the problem for the quadratic function \( f(x) = 4x^2 + 4x - 4 \), we need to find: 1. The vertex 2. The axis of symmetry 3. The x-intercept(s) Step 1: Vertex The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula: \[ x = -\frac{b}{2a} \] Here, \( a = 4 \) and \( b = 4 \). Substitute these values: \[ x = -\frac{4}{2 \times 4} = -\frac{4}{8} = -0.5 \] Now, substitute \( x = -0.5 \) back into the function to find \( y \): \[ f(-0.5) = 4(-0.5)^2 + 4(-0.5) - 4 \] \[ = 4(0.25) - 2 - 4 \] \[ = 1 - 2 - 4 = -5 \] The vertex is \((-0.5, -5)\). Step 2: Axis of Symmetry The axis of symmetry for a quadratic function is the vertical line that passes through the vertex. Therefore, the axis of symmetry is: \[ x = -0.5 \] Step 3: X-intercept(s) To find the x-intercepts, set \( f(x) \) to zero and solve for \( x \): \[ 4x^2 + 4x - 4 = 0 \] Divide the entire equation by 4: \[ x^2 + x - 1 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \). \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -1}}{2 \times 1} \] \[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} \] \[ x = \frac{-1 \pm \sqrt{5}}{2} \] The x-intercepts are \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \). So, the x-intercepts are approximately: \( \left(\frac{-1 + \sqrt{5}}{2}, 0\right) \) and \( \left(\frac{-1 - \sqrt{5}}{2}, 0\right) \). Final Answers: - Vertex: \((-0.5, -5)\) - Axis of Symmetry: \(x = -0.5\) - X-intercepts: \(\left(\frac{-1 + \sqrt{5}}{2}, 0\right), \left(\frac{-1 - \sqrt{5}}{2}, 0\right)\)