Problem 18 - 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. $$f(x)=-3 x^{2}+6 x-2$$

Step-by-Step Solution

Answer

The vertex is at (1, 1), x-intercepts are \( \left(1 \pm \frac{\sqrt{15}}{3}, 0 \right) \), and the y-intercept is (0, -2).

Step 1: Express the Quadratic Function in Standard Form

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

Step 2: Find the Vertex of the Quadratic Function

The vertex of a quadratic function \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting our values, \( x = -\frac{6}{2(-3)} = 1 \). To find the \( y \)-coordinate, substitute \( x = 1 \) back into the function: \( f(1) = -3(1)^2 + 6(1) - 2 = 1 \). So, the vertex is \( (1, 1) \).

Step 3: Find the x-intercepts

To find the \(x\)-intercepts, set \( f(x) = 0 \) and solve the equation \( -3x^2 + 6x - 2 = 0 \). Use the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \). Here, \( a = -3 \), \( b = 6 \), \( c = -2 \). Calculate the discriminant: \( b^2 - 4ac = 36 + 24 = 60 \). Thus, the roots are \( x = \frac{6 \pm \sqrt{60}}{-6} = 1 \pm \frac{\sqrt{15}}{3} \). The \( x \)-intercepts are \( \left(1 + \frac{\sqrt{15}}{3}, 0\right) \) and \( \left(1 - \frac{\sqrt{15}}{3}, 0\right) \).

Step 4: Find the y-intercept

The \( y \)-intercept is where \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = -3(0)^2 + 6(0) - 2 = -2 \). Thus, the \( y \)-intercept is \( (0, -2) \).

Step 5: Sketch the Graph

Using the vertex \((1, 1)\), \(x\)-intercepts \( \left(1 + \frac{\sqrt{15}}{3}, 0\right) \) and \( \left(1 - \frac{\sqrt{15}}{3}, 0\right) \), and \(y\)-intercept \((0, -2)\), sketch a downward-opening parabola. The axis of symmetry is \(x=1\), and the parabola passes through the points identified in the previous steps.

Key Concepts

Vertex Form

The **vertex form** of a quadratic function is a way of expressing the function that makes it easy to identify the vertex. It is given as \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola.
To convert a quadratic function from standard form to vertex form, we need to complete the square. This can help in understanding the shape and position of a parabola quickly.
In the exercise, the vertex was directly calculated by using the formula for the vertex's x-coordinate: \( x = -\frac{b}{2a} \). After finding \( x = 1 \), substituting back into the function gave us the y-coordinate of 1. Therefore, the vertex is \((1, 1)\).
Understanding vertex form is essential because it allows you to see how the graph shifts and transforms from the parent function \( y = x^2 \). It also directly reveals the vertex of the parabola, which is crucial when graphing or converting between forms.

X-Intercepts

Finding the **x-intercepts** of a quadratic function involves setting the function equal to zero and solving for \(x\). This is where the graph of the function crosses the x-axis.
For the quadratic \( f(x) = -3x^2 + 6x - 2 \), we set the equation \( -3x^2 + 6x - 2 = 0 \). The quadratic formula, \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), is a reliable tool to find the x-intercepts for any quadratic equation.
In our exercise, applying this formula, we found the x-intercepts to be \( \left(1 + \frac{\sqrt{15}}{3}, 0\right) \) and \( \left(1 - \frac{\sqrt{15}}{3}, 0\right) \).
These intercepts tell us where the parabola hits the x-axis and indicate the points where the function value is zero. Finding them is crucial in graphing to ensure accuracy.

Y-Intercepts

The **y-intercept** of a quadratic function is the point where the graph intersects the y-axis. This occurs when \(x = 0\).
To find the y-intercept, we substitute 0 for \(x\) in the given quadratic equation. For the function \( f(x) = -3x^2 + 6x - 2 \), substituting \( x = 0 \) gives us \( f(0) = -2 \). Thus, the y-intercept is the point \((0, -2)\).
This point is key when plotting the graph, as it provides a definite position of the parabola on the y-axis. The y-intercept is a straightforward calculation and crucial for the overall shape and understanding of the graph's vertical position.

Parabola Graphing

**Parabola graphing** involves plotting a quadratic function and understanding its shape based on its vertex, intercepts, and symmetry.
To graph the function \( f(x) = -3x^2 + 6x - 2 \):

• Identify the vertex at \((1, 1)\).
• Note the x-intercepts at \( \left(1 + \frac{\sqrt{15}}{3}, 0\right) \) and \( \left(1 - \frac{\sqrt{15}}{3}, 0\right) \).
• Mark the y-intercept at \((0, -2)\).

The axis of symmetry is \(x = 1\), splitting the parabola into mirror images. Because the leading coefficient \(a = -3\) is negative, the parabola opens downwards.
Plotting these critical points and drawing the parabola through them provides a visual representation of the function. It's essential to consider the concavity and symmetry while sketching, which are influenced by the vertex and x-intercepts.