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Chapter 2: Problem 13

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

Short Answer

Expert verified
Vertex: (-1, 1), y-intercept: (0, 3), no real x-intercepts.

Step by step solution

01

Identify Standard Form

The standard form of a quadratic function is \( ax^2 + bx + c \). We are already given the equation in this form: \( f(x) = 2x^2 + 4x + 3 \), with \( a = 2 \), \( b = 4 \), and \( c = 3 \).
02

Find the Vertex

The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the vertex formula \( x = -\frac{b}{2a} \). Substituting \( a = 2 \) and \( b = 4 \) gives: \( x = -\frac{4}{2 \times 2} = -1 \). To find the y-coordinate, substitute \( x = -1 \) back into the function: \( f(-1) = 2(-1)^2 + 4(-1) + 3 = 1 \). Hence, the vertex is \((-1, 1)\).
03

Find the x-intercepts

To find the \( x \)-intercepts, set \( f(x) = 0 \): \( 2x^2 + 4x + 3 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute \( a = 2 \), \( b = 4 \), and \( c = 3 \). Calculate the discriminant: \( b^2 - 4ac = 16 - 24 = -8 \). Since the discriminant is negative, there are no real \( x \)-intercepts.
04

Find the y-intercept

The y-intercept of a function can be found by evaluating the function at \( x=0 \). Substituting into the function, \( f(0) = 2(0)^2 + 4(0) + 3 = 3 \). Thus, the y-intercept is \((0, 3)\).
05

Sketch the Graph

Given that the parabola opens upwards (since \( a = 2 > 0 \)) and its vertex is \((-1, 1)\). The y-intercept is \((0, 3)\), and there are no real \( x \)-intercepts because of the negative discriminant. Plotting these key points provides a sketch of the parabola.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex Formula
The vertex formula is a powerful tool when dealing with quadratic functions. In any quadratic expression of the form \( ax^2 + bx + c \), the vertex is a significant point that shows the maximum or minimum value of the function depending on the direction the parabola opens. To find the vertex, we use the formula:
  • \( x = -\frac{b}{2a} \)
This formula helps us identify the x-coordinate of the vertex. Plug this value back into the original function to obtain the y-coordinate, fully defining the vertex coordinates as \((x, f(x))\). For example, in our problem, substituting \( a = 2 \) and \( b = 4 \) gives \( x = -1 \). Evaluating \( f(-1) = 1 \), hence the vertex is \((-1, 1)\).
Quadratic Formula
When it comes to finding the roots of a quadratic equation, the quadratic formula is a go-to method, especially when the quadratic cannot be easily factored. The formula is given as:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This equation provides the solutions or x-intercepts of the quadratic function. The elements under the square root symbol, \( b^2 - 4ac \), play a crucial role in determining the nature of the roots. In cases where this value (known as the discriminant) is negative, as in our exercise, the quadratic has no real roots, indicating the parabola does not cross the x-axis. In this case, the calculated discriminant was \(-8\).
Discriminant
The discriminant of a quadratic equation \( ax^2 + bx + c \) is an expression defined as \( b^2 - 4ac \). It gives us insights into the nature of the roots of the quadratic equation without actually solving it. The discriminant provides the following indications:
  • If \( b^2 - 4ac > 0 \), there are two distinct real roots.
  • If \( b^2 - 4ac = 0 \), there is one real root (or a repeated root).
  • If \( b^2 - 4ac < 0 \), there are no real roots, only complex ones.
In the given function, the discriminant was calculated as \(-8\), which clearly signifies the absence of real roots. This means the parabola lies entirely above the x-axis, as confirmed by our vertex and other calculation results.
Parabola Sketching
Sketching the graph of a quadratic function provides a visual understanding of its behavior. Key points to consider when sketching a parabola include its vertex, y-intercept, and the nature of its x-intercepts. Let's summarize how to sketch a parabola from the quadratic function:
  • **Direction of Opening:** Defined by the coefficient \( a \). If \( a > 0 \), opens upwards; if \( a < 0 \), opens downwards.
  • **Vertex:** The starting point of our sketch, here at \((-1, 1)\).
  • **Y-intercept:** Found by substituting \( x = 0 \), yielded the point \((0, 3)\).
  • **X-intercepts:** Check using the discriminant; here, it is negative, indicating no real x-intercepts.
With these points, begin with the vertex, plot the y-intercept, and draw a smooth curve opening upwards. This methodical approach helps accurately represent the quadratic function鈥檚 graph.

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