Questions: Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the axis of symmetry. Use the graph to determine the domain and range of the function. f(x)=(x-2)^2-1

Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the axis of symmetry. Use the graph to determine the domain and range of the function.
f(x)=(x-2)^2-1

Transcript text: Use the vertex and intercepts to sketch the graph of the quadratic function. Give the equation of the axis of symmetry. Use the graph to determine the domain and range of the function. \[ f(x)=(x-2)^{2}-1 \]

Solution

Solution Steps

Step 1: Identify the Vertex

The given quadratic function is \( f(x) = (x - 2)^2 - 1 \). This is in the vertex form \( f(x) = a(x - h)^2 + k \), where the vertex is \((h, k)\). Here, \( h = 2 \) and \( k = -1 \). Therefore, the vertex is \((2, -1)\).

Step 2: Determine the Axis of Symmetry

The axis of symmetry for a quadratic function in vertex form \( f(x) = a(x - h)^2 + k \) is the vertical line \( x = h \). For the given function, the axis of symmetry is \( x = 2 \).

Step 3: Find the Intercepts

Y-intercept: Set \( x = 0 \) and solve for \( f(x) \): \[ f(0) = (0 - 2)^2 - 1 = 4 - 1 = 3 \] So, the y-intercept is \((0, 3)\).
X-intercepts: Set \( f(x) = 0 \) and solve for \( x \): \[ 0 = (x - 2)^2 - 1 \implies (x - 2)^2 = 1 \implies x - 2 = \pm 1 \] \[ x = 2 \pm 1 \implies x = 3 \text{ or } x = 1 \] So, the x-intercepts are \((3, 0)\) and \((1, 0)\).