Problem 27 - Precalculus Mathematics for Calculus

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$f(x)=-x^{2}-3 x+3$$

Step-by-Step Solution

Answer

(a) The function is already in standard form. (b) Sketch: vertex at \(\left(\frac{3}{2}, -\frac{15}{4}\right)\), intercepts at y = 3, x via quadratic formula. (c) Maximum is \(-\frac{15}{4}\).

Step 1: Identify quadratic function form

The given quadratic function is in the standard quadratic form, which is \( f(x) = ax^2 + bx + c \). Here, \( a = -1 \), \( b = -3 \), and \( c = 3 \). The function is already in the standard form.

Step 2: Determine the vertex of the parabola

To find the vertex, use the formula \( x = -\frac{b}{2a} \). Substituting the values, we get \( x = -\frac{-3}{2(-1)} = \frac{3}{2} \). To find the y-coordinate of the vertex, plug \( x = \frac{3}{2} \) back into the function: \( f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 3 = -\frac{9}{4} - \frac{9}{2} + 3 = -\frac{9}{4} - \frac{18}{4} + \frac{12}{4} = -\frac{15}{4} \). So, the vertex is \( \left(\frac{3}{2}, -\frac{15}{4}\right) \).

Step 3: Sketch the graph

Since \( a = -1 \), the parabola opens downwards. The vertex \( \left(\frac{3}{2}, -\frac{15}{4}\right) \) is the maximum point of the parabola. To sketch, plot the vertex and note that it is a downward-opening parabola. Use additional points (such as intercepts) to help draw an accurate curve.

Step 4: Find intercepts to assist in sketching

The y-intercept occurs when \( x = 0 \). Substituting into the function: \( f(0) = -0^2 - 3(0) + 3 = 3 \). So, the y-intercept is \( (0, 3) \). For x-intercepts, set \( f(x) = 0 \) and solve \(-x^2 - 3x + 3 = 0 \), which can be solved using the quadratic formula. Calculate discriminant: \( b^2 - 4ac = (-3)^2 - 4(-1)(3) = 9 + 12 = 21 \). Since the discriminant is positive, there are two real intercepts at \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 5: Calculate x-intercepts using Quadratic formula

Using the quadratic formula: \( x = \frac{-(-3) \pm \sqrt{21}}{2(-1)} = \frac{3 \pm \sqrt{21}}{-2} \). This yields two solutions: \( x = \frac{3 + \sqrt{21}}{-2} \) and \( x = \frac{3 - \sqrt{21}}{-2} \). These provide the x-intercepts for graph sketching.

Step 6: Identify maximum or minimum value

Since the parabola opens downwards, the vertex represents the maximum point of the function. Thus, the maximum value of \( f(x) \) is \( -\frac{15}{4} \), occurring at \( x = \frac{3}{2} \).

Key Concepts

Standard Form of a Quadratic Function

The standard form of a quadratic function is expressed as \( f(x) = ax^2 + bx + c \). This is the most straightforward and widely-used representation of quadratic functions. Here,

• \( a \), \( b \), and \( c \) are coefficients with \( a eq 0 \).
• \( a \) determines the direction the parabola opens. If \( a > 0 \), it opens upwards. If \( a < 0 \), it opens downwards.
• \( b \) influences the parabola’s axis of symmetry and vertex position.
• \( c \) is the y-intercept of the graph, where the parabola crosses the y-axis.

Given the function \( f(x) = -x^2 - 3x + 3 \), we identify it as already being in standard form with \( a = -1 \), \( b = -3 \), and \( c = 3 \). This sets the stage for analyzing the shape and properties of the parabola such as finding its vertex and determining if it reaches a maximum or minimum value.

Vertex of a Parabola

The vertex of a parabola provides a critical insight into its graph, acting as either a peak or a trough point, depending on the parabola's direction. The formula used to find the vertex, specifically its x-coordinate, is \( x = -\frac{b}{2a} \).
For our function \( f(x) = -x^2 - 3x + 3 \), substituting our known values gives: \[ x = -\frac{-3}{2(-1)} = \frac{3}{2} \].
The y-coordinate of the vertex on the graph can be found by plugging the x-value into the function: \[ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 3 \].
After simplifying, this results in a y-value of \(-\frac{15}{4}\). Therefore, the vertex of this parabola is at \( \left(\frac{3}{2}, -\frac{15}{4}\right) \), indicating where the graph reaches its maximum or minimum point.
In this case, with \( a = -1 \), the parabola opens downward, and hence, the vertex represents the maximum point of the function's output.

Maximum and Minimum Values

The maximum or minimum value of a quadratic function is directly related to the vertex and the direction the parabola opens. For downward opening parabolas, like \( f(x) = -x^2 - 3x + 3 \) where \( a < 0 \), the vertex is the maximum point. Conversely, if the parabola opened upward (\( a > 0 \)), the vertex would be the minimum.
Using the previously calculated vertex coordinates, \( \left(\frac{3}{2}, -\frac{15}{4}\right) \), the maximum value of this function is \(-\frac{15}{4}\). This value occurs at the x-coordinate this vertex represents, \( x = \frac{3}{2} \).
To summarize:

• If \( a > 0 \): The vertex is the lowest point, giving the minimum value.
• If \( a < 0 \): The vertex becomes the highest point, giving the maximum value.

Finding these values is essential in various applications such as determining optimal solutions in real-world quadratic problems.