Problem 16 - Precalculus Mathematics for Calculus

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x\) - and \(y\) -intercept(s). (c) Sketch its graph. $$f(x)=-x^{2}-4 x+4$$

Step-by-Step Solution

Answer

The function is in standard form already. The vertex is \((2, -8)\), \( x \)-intercepts are \((-2 + 2\sqrt{2}, 0)\) and \((-2 - 2\sqrt{2}, 0)\), and \( y \)-intercept is \((0, 4)\).

Step 1: Express in Standard Form

A quadratic function in standard form is written as \( f(x) = ax^2 + bx + c \). Here, the given function is already in standard form with \( a = -1 \), \( b = -4 \), and \( c = 4 \).

Step 2: Find the Vertex

To find the vertex of the quadratic function, use the formula \( x = -\frac{b}{2a} \). Here, \( b = -4 \) and \( a = -1 \). Calculate: \( x = -\frac{-4}{2 \, (-1)} = 2 \). Substitute \( x = 2 \) back into the function to find \( y \): \( f(2) = -2^2 - 4 \times 2 + 4 = -4 - 8 + 4 = -8 \). Thus, the vertex is \((2, -8)\).

Step 3: Find the x-Intercepts

Set \( f(x) = 0 \) and solve the equation \( -x^2 - 4x + 4 = 0 \). Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( b = -4 \), \( a = -1 \), and \( c = 4 \). Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4 \times (-1) \times 4 = 16 + 16 = 32 \). Then, find \( x \): \( x = \frac{4 \pm \sqrt{32}}{-2} \). Simplify \( \sqrt{32} = 4\sqrt{2} \). Thus, \( x = \frac{4 \pm 4\sqrt{2}}{-2} \), giving two solutions: \( x = -2 + 2\sqrt{2} \) and \( x = -2 - 2\sqrt{2} \). These are the \( x \)-intercepts.

Step 4: Find the y-Intercept

The \( y \)-intercept is found by setting \( x = 0 \) in the function: \( f(0) = -(0)^2 - 4 \times 0 + 4 = 4 \). Thus, the \( y \)-intercept is \( (0, 4) \).

Step 5: Sketch the Graph

Using the vertex \( (2, -8) \), \( x \)-intercepts \( (-2 + 2\sqrt{2}, 0) \) and \( (-2 - 2\sqrt{2}, 0) \), and the \( y \)-intercept \( (0, 4) \), sketch the parabola. Since \( a = -1 \), the parabola opens downwards. Plot these key points and draw a smooth curve passing through them.

Key Concepts

Vertex

The vertex of a quadratic function is a crucial point that indicates the peak or the trough of the parabola. For any quadratic function in standard form, the vertex can be located using the formula \( x = -\frac{b}{2a} \). Plug the value of \( x \) into the function to find the corresponding \( y \) value. This gives you the vertex, expressed as a coordinate \((x, y)\).
For the function \( f(x) = -x^2 - 4x + 4 \), we calculate the \( x \)-value of the vertex as 2, leading to a \( y \)-value of -8 once substituted back. Thus, this function's vertex is \((2, -8)\).
The vertex is a vital point as it reflects the function's maximum or minimum value in a specific interval. Here, since the parabola opens downwards (the coefficient of \( x^2 \) is negative), the vertex is a maximum point.

x-intercepts

The \( x \)-intercepts of a quadratic function are the points where the graph intersects the \( x \)-axis. These occur where the function equals zero, \( f(x) = 0 \). You can find these intercepts by solving the quadratic equation using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For \( f(x) = -x^2 - 4x + 4 \), this results in two potential \( x \)-values: \(-2 + 2\sqrt{2}\) and \(-2 - 2\sqrt{2}\).
These values represent where the parabola cuts the \( x \)-axis, providing critical points for accurately sketching the graph. Understanding \( x \)-intercepts also helps in analyzing the function's real roots.

y-intercepts

A quadratic function's \( y \)-intercept is the point where the graph intersects the \( y \)-axis. This can be determined by evaluating the quadratic function at \( x = 0 \).

• Plug in \( x = 0 \) within the given equation \( f(x) = -x^2 - 4x + 4 \), resulting in the \( y \)-intercept \( f(0) = 4 \).

Thus, the \( y \)-intercept is \((0, 4)\).
The \( y \)-intercept gives a starting point for sketching the parabola and represents the output of the quadratic when no input (or zero input) is provided.

Standard Form

The standard form of a quadratic function is useful as it provides quick access to information necessary for graphing. It is expressed as \( f(x) = ax^2 + bx + c \). The coefficients \( a \), \( b \), and \( c \) directly show the direction and shape of the parabola:

• \( a \): If positive, the parabola opens upwards; if negative, it opens downwards.
• \( b \): Influences the horizontal placement of the vertex.
• \( c \): Indicates the \( y \)-intercept directly.

For the function \( f(x) = -x^2 - 4x + 4 \), it is already in standard form with \( a = -1 \), \( b = -4 \), and \( c = 4 \).
This facilitates finding all other important features of the parabola like vertex and intercepts, which are crucial for analyzing and sketching the function.

Graph Sketching

Graph sketching becomes simpler by identifying key characteristics such as vertex, intercepts, and the direction of the parabola (upwards or downwards). Begin by plotting the vertex which serves as the core structural point of the parabola. Next, plot the \( x \)-intercepts and \( y \)-intercept:

• Vertex: \((2, -8)\)
• \( x \)-intercepts: \((-2 + 2\sqrt{2}, 0)\) and \((-2 - 2\sqrt{2}, 0)\)
• \( y \)-intercept: \((0, 4)\)

Connect these points with a smooth curve. Since \( a = -1 \), the parabola opens downwards, indicating that it is an upside-down "U" shape.
Visually capturing these features will help in better understanding the quadratic function's behavior, allowing an accurate sketch of the parabolic graph.