Question

(a) find the vertex and the axis of symmetry of each quadratic function, and determine whether the graph is concave up or concave down. (b) Find the y-intercept and the \(x\) -intercepts, if any. (c) Use parts (a) and (b) to graph the function. (d) Find the domain and the range of the quadratic function. (e) Determine where the quadratic function is increasing and where it is decreasing. (f) Determine where \(f(x)>0\) and where \(f(x)<0\) \(f(x)=-x^{2}+4 x\)

Short answer

Vertex: (2, 4). Axis: x = 2. Concave down. y-int: (0, 0). x-ints: (0, 0) and (4, 0). Domain: \((-\infty, \infty)\). Range: \((-\infty, 4]\). Increase: \( ( -\infty, 2)\) Decrease: \((2,\infty)\).\( f(x) > 0, x \in (0, 4)\), \( f(x) < 0, x < 0 \) or \( x > 4 \).

Step-by-step solution

- Find the Vertex

The standard form of a quadratic function is given by \[f(x) = ax^2 + bx + c.\]Given the function \(f(x) = -x^2 + 4x\), we identify \(a = -1\), \(b = 4\), and \(c = 0\). The vertex formula is: \[x = -\frac{b}{2a}.\]Substituting the values of \(a\) and \(b\), we get \[x = -\frac{4}{2(-1)} = 2.\] Now, to find the \(y\)-coordinate of the vertex, substitute \(x = 2\) back into the function: \[f(2) = -(2)^2 + 4(2) = -4 + 8 = 4.\] So, the vertex is \((2, 4)\).

- Determine the Axis of Symmetry

The axis of symmetry for a quadratic function is always the vertical line passing through the vertex's \(x\)-coordinate. From Step 1, the vertex is \((2, 4)\), so the axis of symmetry is \[x = 2.\]

- Determine Concavity

To determine whether the graph is concave up or concave down, look at the coefficient \(a\). Since \(a = -1\), which is less than 0, the graph is concave down.

- Find the Y-Intercept

The \(y\)-intercept is found by setting \(x = 0\) and solving for \(f(x)\): \[f(0) = -(0)^2 + 4(0) = 0.\]So, the \(y\)-intercept is \((0, 0)\).

- Find the X-Intercepts

The \(x\)-intercepts are found by setting \(f(x) = 0\) and solving the equation:\[-x^2 + 4x = 0,\]which factors to \[-x(x - 4) = 0.\]Solving for \(x\), we get \(x = 0\) and \(x = 4\). So, the \(x\)-intercepts are \((0, 0)\) and \((4, 0)\).

- Graph the Function

Use the vertex \((2, 4)\), axis of symmetry \(x = 2\), and the intercepts \((0, 0)\) and \((4, 0)\) to sketch the graph. Plot these points and draw a parabola opening downwards.

- Domain and Range

The domain of any quadratic function is all real numbers: \[(-\infty, \infty).\]The range of a concave down parabola with vertex \((2, 4)\) is \[(-\infty, 4].\]

- Increasing and Decreasing Intervals

A concave down parabola increases to the left of the vertex and decreases to the right. Therefore, increasing on the interval: \[ ( -\infty, 2)\] and decreasing on the interval: \[(2,\infty).\]

- Where \(f(x) > 0\) and \(f(x) < 0\)

The function \( f(x) > 0 \) between the \(x\)-intercepts \( (0, 4) \). The function \( f(x) < 0 \) for \(x < 0\) and \(x > 4.\)

Key concepts

vertex

The vertex of a quadratic function is the highest or lowest point on its graph. It determines the maximum or minimum value of the function. To find the vertex, we use the formula: \[x = -\frac{b}{2a}\]. For the function \(f(x) = -x^2 + 4x\), \(a = -1\) and \(b = 4\). Plugging in these values, we get \(x = 2\). Substituting \(x = 2\) back into the function, we find the y-coordinate: \[f(2) = -(2)^2 + 4(2) = 4\]. Hence, the vertex is \((2, 4)\).

axis of symmetry

The axis of symmetry of a quadratic function is a vertical line that passes through the vertex, acting as a mirror line for the parabola. For the function \(f(x) = -x^2 + 4x\), the vertex is \((2, 4)\). Thus, the axis of symmetry is the line \(x = 2\). This line divides the parabola into two symmetrical halves.

concavity

The concavity of a parabola tells us whether it curves upwards or downwards. It is determined by the coefficient \(a\) in the quadratic equation \(ax^2 + bx + c\). If \(a > 0\), the parabola is concave up (U-shaped). If \(a < 0\), it is concave down (n-shaped). For the function \(f(x) = -x^2 + 4x\), \(a = -1\), which is less than 0, so the graph is concave down.

y-intercept

The y-intercept is where the graph crosses the y-axis, found by setting \(x = 0\). For the function \(f(x) = -x^2 + 4x\), we substitute \(x = 0\): \[f(0) = -(0)^2 + 4(0) = 0\]. Therefore, the y-intercept is \((0, 0)\).

x-intercepts

The x-intercepts are the points where the graph crosses the x-axis. This happens when \(f(x) = 0\). For \(f(x) = -x^2 + 4x\), we solve: \[-x^2 + 4x = 0\], which factors to \[-x(x - 4) = 0\]. Solving for \(x\), we get \(x = 0\) and \(x = 4\), so the x-intercepts are \((0, 0)\) and \((4, 0)\).

domain

The domain of a quadratic function is the set of all possible input values (x-values). For any quadratic function of the form \(ax^2 + bx + c\), the domain is all real numbers: \((-\infty, \infty)\). This means you can substitute any real number for \(x\) in the function.

range

The range of a quadratic function is the set of possible output values (y-values). For a concave down parabola like \(f(x) = -x^2 + 4x\) with a vertex at \((2, 4)\), the range is \((-\infty, 4]\). This means the highest value is 4, and the y-values go downwards infinitely.

increasing intervals

A quadratic function increases to a point (the vertex) and then decreases. For the function \(f(x) = -x^2 + 4x\), the vertex is at \((2, 4)\). The function increases on the interval \((-\infty, 2)\) as you approach the vertex from the left side.

decreasing intervals

Similarly, a quadratic function decreases after reaching the vertex. For the function \(f(x) = -x^2 + 4x\), the intervals where the function decreases are to the right of the vertex. Therefore, it decreases on the interval \((2, \infty)\).

quadratic inequality

Quadratic inequalities determine where the function is greater or less than zero. For \(f(x) = -x^2 + 4x\), the function \(f(x) > 0\) between the x-intercepts \(0\) and \(4\) (i.e., \((0, 4)\)). The function \(f(x) < 0\) for \(x < 0\) and \(x > 4\), outside these intercepts.