Problem 17 - 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. $$ f(x)=2 x^{2}+4 x+3 $$

Step-by-Step Solution

Answer

The quadratic is in standard form; vertex \((-1, 1)\); y-intercept \((0, 3)\); no real x-intercepts.

Step 1: Convert to Standard Form

The given quadratic function is in the form \(f(x) = ax^2 + bx + c\). In this case, the function is already in standard form: \(f(x) = 2x^2 + 4x + 3\). Therefore, no conversion is necessary.

Step 2: Find the Vertex

The vertex of a quadratic function \(ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 2\) and \(b = 4\).Calculate:\[x = -\frac{4}{2 imes 2} = -1\]Substitute \(x = -1\) back into the function to find \(y\):\[f(-1) = 2(-1)^2 + 4(-1) + 3 = 2 - 4 + 3 = 1\]So, the vertex is \((-1, 1)\).

Step 3: Find the x-intercepts

The x-intercepts occur where \(f(x) = 0\). Set the quadratic equation equal to zero:\[2x^2 + 4x + 3 = 0\]Use the quadratic formula where \(a = 2\), \(b = 4\), and \(c = 3\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Calculate:\[x = \frac{-4 \pm \sqrt{4^2 - 4(2)(3)}}{2(2)} = \frac{-4 \pm \sqrt{16 - 24}}{4} = \frac{-4 \pm \sqrt{-8}}{4}\]Since the discriminant \(-8\) is negative, there are no real x-intercepts.

Step 4: Find the y-intercept

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function:\[f(0) = 2(0)^2 + 4(0) + 3 = 3\]So, the y-intercept is \((0, 3)\).

Step 5: Sketch the Graph

To graph \(f(x) = 2x^2 + 4x + 3\), note the following:- **Vertex**: \((-1, 1)\)- **y-intercept**: \((0, 3)\)- **No x-intercepts**: Since the parabola does not cross the x-axis due to a negative discriminant.The parabola opens upwards (since \(a = 2 > 0\)) and is shifted according to its vertex. Plot the points and draw a smooth, symmetric curve through the vertex and y-intercept.

Key Concepts

Standard Form

In the context of quadratic functions, the standard form is vital because it defines the structure and key elements of the function. A quadratic function in standard form is expressed as:

• \(f(x) = ax^2 + bx + c\)

Here, \(a\), \(b\), and \(c\) are real numbers and \(a eq 0\). The coefficients \(a\), \(b\), and \(c\) give insight into the parabola's shape and position. In our specific problem, the quadratic function is \(f(x) = 2x^2 + 4x + 3\). This means we are already given the function in the standard form. Recognizing the standard form allows us to easily apply formulas for finding the vertex, x-intercepts, and y-intercepts. The value of \(a\) informs us whether the parabola opens upwards or downwards. In this case, since \(a = 2\) which is greater than zero, the parabola opens upwards.

Vertex Formula

The vertex of a parabola is a crucial point, often representing its peak or trough, depending on the orientation. The vertex formula provides an efficient method to determine this point using the function's coefficients. For a quadratic function \(ax^2 + bx + c\), the x-coordinate of the vertex is calculated using:

• \(x = -\frac{b}{2a}\)

Substituting \(a = 2\) and \(b = 4\) into the formula gives:

• \(x = -\frac{4}{2 \times 2} = -1\)

The y-coordinate is found by plugging \(x = -1\) back into the original function:

• \(f(-1) = 2(-1)^2 + 4(-1) + 3 = 1\)

Thus, the vertex of the parabola is at \((-1, 1)\). This vertex tells us that the graph reaches its minimum value at this point due to the upward opening of the parabola.

X-Intercepts

X-intercepts are points where the graph of the function crosses the x-axis. For quadratic functions, x-intercepts are found by setting the quadratic equation equal to zero and solving for \(x\). In our specific function, we solve:

• \(2x^2 + 4x + 3 = 0\)

We apply the quadratic formula:

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

With \(a = 2\), \(b = 4\), and \(c = 3\), the discriminant \(b^2 - 4ac = 16 - 24 = -8\) is negative.

• This results in no real number solutions for \(x\), indicating the parabola does not cross the x-axis and there are no real x-intercepts.

A negative discriminant is a simple yet powerful indication that the graph sits completely above or below the x-axis.

Y-Intercepts

The y-intercept is a point where the graph of the function crosses the y-axis. It is obtained by setting \(x = 0\) and calculating \(f(0)\). In our problem, substituting gives:

• \(f(0) = 2(0)^2 + 4(0) + 3 = 3\)

The y-intercept is thus \((0, 3)\). This simple calculation provides crucial information on where the parabola meets the y-axis. Having this point helps when plotting the overall graph of the function, serving as a starting point for drawing.

Graphing Parabolas

Graphing a quadratic function involves plotting key points, like the vertex and intercepts, and understanding the parabola's direction. Here's a step-by-step guide for our quadratic function:

• Identify the vertex: It is \((-1, 1)\).
• Determine the intercepts: Y-intercept is \((0, 3)\).
• No x-intercepts: Because the discriminant is negative.
• Determine the parabola's direction: The parabola opens upwards since \(a = 2 > 0\).

After identifying these points and direction, we connect them with a smooth curve. It is important that the curve is symmetrical around the vertex, reflecting the fundamental property of parabolas. Always check to ensure the parabola aligns with the intercepts and maintains the correct direction of opening. This visual representation greatly enhances understanding of the function's behavior.