Problem 18 - Algebra and Trigonometry

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. $$ f(x)=-3 x^{2}+6 x-2 $$

Step-by-Step Solution

Answer

Vertex: (1, 1); x-intercepts: \(\frac{1+\sqrt{3}}{3}, \frac{1-\sqrt{3}}{3}\); y-intercept: (0, -2).

Step 1: Express in Standard Form

The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). The given quadratic function is already in the standard form: \( f(x) = -3x^2 + 6x - 2 \).

Step 2: Find the Vertex

The vertex of a quadratic function given in standard form \( ax^2 + bx + c \) is \( \left( \frac{-b}{2a}, f \left( \frac{-b}{2a} \right) \right) \). Here, \( a = -3 \), \( b = 6 \). So, calculate \( x = \frac{-6}{2(-3)} = 1 \). Then find the vertex by substituting \( x = 1 \) into \( f(x) \): \( f(1) = -3(1)^2 + 6(1) - 2 = 1 \). Hence, the vertex is \((1, 1)\).

Step 3: Find the x-intercepts

To find the x-intercepts, solve \( -3x^2 + 6x - 2 = 0 \). Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -3 \), \( b = 6 \), \( c = -2 \). Calculate the discriminant: \( b^2 - 4ac = 6^2 - 4(-3)(-2) = 12 \). Thus, \( x = \frac{-6 \pm \sqrt{12}}{-6} = \frac{-6 \pm 2\sqrt{3}}{-6} = \frac{1 \pm \sqrt{3}}{3} \). So, the x-intercepts are \( \frac{1+\sqrt{3}}{3}, \frac{1-\sqrt{3}}{3} \).

Step 4: Find the y-intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into \( f(x) \): \( f(0) = -3(0)^2 + 6(0) - 2 = -2 \). Hence, the y-intercept is \((0, -2)\).

Step 5: Sketch the Graph

Using the previously calculated vertex, x-intercepts, and y-intercept, draw the graph. The vertex \((1, 1)\), x-intercepts \(\left( \frac{1+\sqrt{3}}{3}, 0 \right), \left( \frac{1-\sqrt{3}}{3}, 0 \right)\), and y-intercept \((0, -2)\) help in sketching. The parabola opens downwards as the coefficient of \(x^2\) is negative.

Key Concepts

Vertex of a Parabola

In a quadratic function, the vertex represents a crucial point where the parabola either reaches its peak or its lowest point. For the function given in the exercise, \( f(x) = -3x^2 + 6x - 2 \), the vertex helps us understand the behavior and direction of the parabola. The formula for finding the vertex \( (h, k) \) in the standard form \( ax^2 + bx + c \) is \( \left( \frac{-b}{2a}, f\left(\frac{-b}{2a}\right) \right) \). Here, the vertex coordinates can be determined by using our specific values of \( a = -3 \) and \( b = 6 \):

• The x-coordinate of the vertex is \( \frac{-6}{2(-3)} = 1 \).
• Substitute this into \( f(x) \) to find the y-coordinate: \( f(1) = 1 \).

Thus, the vertex of the parabola is \((1, 1)\). This point represents the maximum for this particular quadratic as \( a \) is negative, which means the parabola opens downward.

X-Intercepts

The x-intercepts of a quadratic function are where the graph crosses the x-axis, meaning \( f(x) = 0 \). To find them, you solve the equation \( -3x^2 + 6x - 2 = 0 \). This quadratic equation can be tackled using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \( a = -3 \), \( b = 6 \), and \( c = -2 \).

• First, calculate the discriminant: \( b^2 - 4ac = 12 \).
• Use this in the quadratic formula to find the roots: \( x = \frac{-6 \pm 2\sqrt{3}}{-6} \).
• This simplifies to the x-intercepts: \( \frac{1+\sqrt{3}}{3} \) and \( \frac{1-\sqrt{3}}{3} \).

These values are where the parabola crosses the x-axis, providing insights into the function's roots or solutions.

Y-Intercept

Finding the y-intercept of a quadratic function reveals where the parabola crosses the y-axis, setting \( x = 0 \). In our case, for the function \( f(x) = -3x^2 + 6x - 2 \):

• Substitute \( x = 0 \) into the function: \( f(0) = -2 \).

Thus, the y-intercept is the point \( (0, -2) \). This point is vital as it is a straightforward feature that marks where the graph starts crossing the vertical axis of a coordinate plane. It's a great initial checkpoint when plotting the graph of the quadratic function.

Standard Form of a Quadratic Function

The standard form of a quadratic function is essential for identifying key properties of the parabola. It is given by the equation \( f(x) = ax^2 + bx + c \). Each component plays a role:

• \( a \), the coefficient of \( x^2 \), dictates the parabola's direction—opening upwards if positive and downwards if negative.
• \( b \) affects the position of the vertex along the x-axis.
• \( c \) provides the y-intercept directly.

In our exercise, the function \( f(x) = -3x^2 + 6x - 2 \) is already in standard form:

• \( a = -3 \) causes the parabola to open downward.
• The features can be easily analyzed thanks to its simple structure. You can use this form to quickly assess the shape and significant points of the graph.

Graphing Quadratics

Graphing a quadratic function involves plotting several crucial points derived from the function's equation. For the function \( f(x) = -3x^2 + 6x - 2 \), you can follow a straightforward approach.

• Begin with the vertex \((1, 1)\) as it gives a pivotal point around which the parabola is symmetric.
• Next, mark the x-intercepts \( \left( \frac{1+\sqrt{3}}{3}, 0 \right) \) and \( \left( \frac{1-\sqrt{3}}{3}, 0 \right) \), indicative of where the graph crosses the x-axis.
• Plot the y-intercept \((0, -2)\), showcasing where the parabola meets the y-axis.

When connecting these points, remember the curve is symmetric along a vertical line through the vertex. As the leading coefficient \( a = -3 \) signals, the parabola will curve downward, arching towards the x-axis from the vertex. This visualization captures the trajectory of the graph, conveying the overall behavior of the quadratic function.