In Exercises 1–4, use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation for the parabola's axis of symmetry. Use the graph to determine the function's domain and range.
$f\left(x\right)=-(x+1{)}^2+4$

Verified step by step guidance

1. Identify the vertex of the quadratic function given in vertex form \(f(x) = - (x + 1)^2 + 4\). The vertex form is \(f(x) = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Here, rewrite \((x + 1)^2\) as \((x - (-1))^2\), so the vertex is at \((-1, 4)\).
2. Determine the axis of symmetry using the vertex. The axis of symmetry is the vertical line that passes through the vertex, so its equation is \(x = -1\).
3. Find the y-intercept by evaluating \(f(0)\). Substitute \(x = 0\) into the function: \(f(0) = - (0 + 1)^2 + 4\). This gives the point where the graph crosses the y-axis.
4. Find the x-intercepts by setting \(f(x) = 0\) and solving for \(x\): \$0 = - (x + 1)^2 + 4$. Rearrange and solve the resulting equation to find the x-values where the graph crosses the x-axis.
5. Determine the domain and range of the function. The domain of any quadratic function is all real numbers, so \((-\infty, \infty)\). Since the parabola opens downward (because \(a = -1 < 0\)) and the vertex is the maximum point at \(y = 4\), the range is \((-\infty, 4]\).

Key Concepts

Vertex Form of a Quadratic Function

The vertex form of a quadratic function is f(x) = a(x - h)^2 + k, where (h, k) is the vertex of the parabola. It helps easily identify the vertex, which is the highest or lowest point depending on the sign of 'a'. In the given function, f(x) = -(x + 1)^2 + 4, the vertex is (-1, 4).

Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola, dividing it into two mirror-image halves. Its equation is x = h, where h is the x-coordinate of the vertex. For the function f(x) = -(x + 1)^2 + 4, the axis of symmetry is x = -1.

Domain and Range of Quadratic Functions

The domain of any quadratic function is all real numbers since x can take any value. The range depends on the vertex and the direction the parabola opens. For f(x) = -(x + 1)^2 + 4, the parabola opens downward, so the range is all y-values less than or equal to 4.