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A quadratic function is typically expressed in its standard form as \(f(x) = ax^2 + bx + c\). This is an efficient way to represent the function because it provides all the coefficients at a glance:

• \(a\): the coefficient of \(x^2\), which also determines the direction the parabola opens (upward when \(a > 0\) and downward when \(a < 0\)).
• \(b\): the linear coefficient that influences the direction and steepness of the slope.
• \(c\): the constant term which represents the \(y\)-intercept of the quadratic.

In our specific example, the function provided is \(f(x) = -x^2 + 10x\). Here, we notice:

• \(a = -1\), indicating the parabola opens downward.
• \(b = 10\), affecting the slope's steepness.
• \(c = 0\), aligning with the \(y\)-intercept at the origin \((0, 0)\).

The vertex of a quadratic function is a significant point where the parabola either reaches a maximum or minimum value. It is represented in the vertex form of the quadratic equation: \(f(x) = a(x-h)^2 + k\), with the vertex \((h, k)\). Finding the vertex can help in understanding the shape and position of the function's graph.

To find the vertex, calculate \(h\) using the formula \(h = -\frac{b}{2a}\). Substituting the values from the equation \(-x^2 + 10x\), we get:\[ h = -\frac{10}{2(-1)} = 5\]Next, input \(h\) back into the function to find \(k\):\[ k = f(5) = -(5)^2 + 10(5) = -25 + 50 = 25\]Thus, the vertex is at \((5, 25)\), which is the highest point of the parabola due to it opening downward.

The \(x\)-intercepts are points where the graph of the quadratic function crosses the \(x\)-axis. These are found by setting \(f(x) = 0\) and solving for \(x\). For our function \(f(x) = -x^2 + 10x\), solving \(-x^2 + 10x = 0\) involves factoring:\[ -x(x - 10) = 0\]This yields the solutions:

• \(x = 0\)
• \(x = 10\)

Therefore, the \(x\)-intercepts are \((0, 0)\) and \((10, 0)\). These intercepts show where the graph touches the \(x\)-axis.

The \(y\)-intercept of a quadratic function is where the graph intersects the \(y\)-axis. This occurs when \(x = 0\), providing an easy calculation by substituting \(0\) into the function. For our specific function \(f(x) = -x^2 + 10x\), evaluating \(f(0)\) gives:\[ f(0) = -0^2 + 10 \times 0 = 0\]Hence, the \(y\)-intercept is \((0, 0)\), which coincidentally is also one of the \(x\)-intercepts. This point shows where the graph starts on the \(y\)-axis.

Graphing a quadratic function provides a visual interpretation of the algebraic equation. The graph of a quadratic is a parabola, symmetric about a vertical axis known as the axis of symmetry, which goes through the vertex.

For the function \(f(x) = -x^2 + 10x\), start with:

• The vertex at \((5, 25)\).
• The \(x\)-intercepts at \((0, 0)\) and \((10, 0)\).
• The \(y\)-intercept at \((0, 0)\).

Since \(a = -1\), the parabola opens downward. This means the graph will rise to the vertex before falling, forming a symmetric and bell-shaped curve. The axis of symmetry is the vertical line \(x = 5\), bisecting the parabola into two mirror-image halves. These key points guide you in accurately sketching the quadratic's curve.