Problem 7 - Algebra and Trigonometry

Question

\(5-18=\) 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. $$ f(x)=2 x^{2}+6 x $$

Step-by-Step Solution

Answer

The vertex is \((-1.5, -4.5)\), \( x \)-intercepts are 0 and -3, \( y \)-intercept is 0.

Step 1: Identify the Standard Form of a Quadratic Function

The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In \( f(x) = 2x^2 + 6x \), we already have the function in standard form with \( a = 2 \), \( b = 6 \), and \( c = 0 \).

Step 2: Find the Vertex

The vertex \((h, k)\) of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( h = \frac{-b}{2a} \). For our function, \( a = 2 \) and \( b = 6 \), so \( h = \frac{-6}{2 \times 2} = \frac{-6}{4} = -1.5 \). Substitute \( x = -1.5 \) back into the function to find \( k \):\( f(-1.5) = 2(-1.5)^2 + 6(-1.5) = 2(2.25) - 9 = 4.5 - 9 = -4.5 \). Thus, the vertex is \((-1.5, -4.5)\).

Step 3: Find the x-intercepts

The \( x \)-intercepts occur where \( f(x) = 0 \). For the function \( 2x^2 + 6x = 0 \), factor out an \( x \):\( x(2x + 6) = 0 \).This results in two solutions: \( x = 0 \) and \( 2x + 6 = 0 \). Solving for \( x \) in the latter, we have \( 2x = -6 \) so \( x = -3 \). Thus, the \( x \)-intercepts are 0 and -3.

Step 4: Find the y-intercept

The \( y \)-intercept of a function is found by evaluating \( f(x) \) at \( x = 0 \). Substituting \( x = 0 \) into the function: \( f(0) = 2(0)^2 + 6(0) = 0 \). Hence, the \( y \)-intercept is 0.

Step 5: Sketch the Graph

To sketch the graph, plot the vertex at \((-1.5, -4.5)\), the \( x \)-intercepts at \( x = 0 \) and \( x = -3 \), and the \( y \)-intercept at \( y = 0 \). Since the coefficient of \( x^2 \) is positive, the parabola opens upwards. Draw a symmetrical parabolic curve through these points to represent the graph of the function.

Key Concepts

Standard Form

The concept of a quadratic function revolves around its standard form, which is expressed as \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants that define the specific characteristics of the parabola. For the function \( f(x) = 2x^2 + 6x \), it is already in standard form with:

• \( a = 2 \)
• \( b = 6 \)
• \( c = 0 \)

This is important because it sets the ground for further analysis such as finding the vertex and intercepts. Understanding that this function doesn't have a \( c \) term means the graph will intersect the y-axis at the origin. This initial observation can give insights into symmetry properties and simplifies calculations for the intercepts.

Vertex

A key feature of any quadratic function is its vertex. The vertex can be thought of as the peak or the lowest point of the parabola, depending on the orientation. To find the vertex of the quadratic \( f(x) = ax^2 + bx + c \), we use the formulas:

• \( h = \frac{-b}{2a} \)
• Plug \( h \) back into the function to find \( k \)

For our specific function \( f(x) = 2x^2 + 6x \), with \( a = 2 \) and \( b = 6 \):

• Compute \( h = \frac{-6}{2 \times 2} = -1.5 \)
• Substitute \( h \) back to get \( k \): \( f(-1.5) = 2(-1.5)^2 + 6(-1.5) = -4.5 \)

Thus, the vertex is \((-1.5, -4.5)\). The vertex indicates the minimum point of the parabola as \( a \) is positive, which also confirms that the parabola opens upwards.

Intercepts

Intercepts, including x-intercepts and the y-intercept, give us important points where the graph of a function crosses the axes. To find the **x-intercepts**:

• Set \( f(x) = 0 \)
• Solve \( 2x^2 + 6x = 0 \)
• Factor: \( x(2x + 6) = 0 \)
• Solve for \( x \): gives solutions \( x = 0 \) and \( x = -3 \)

These zeros of the function indicate where the graph intersects the x-axis.For the **y-intercept**:

• Evaluate \( f(x) \) at \( x = 0 \)
• This results in \( f(0) = 0 \)

Thus, the y-intercept is \( (0, 0) \). Together, these intercepts help sketch out essential structure points of the graph.

Graph of a Function

The graph of a quadratic function like \( f(x) = 2x^2 + 6x \) gives a visual representation of all its characteristics.To graph it:

• Plot the **vertex**: \((-1.5, -4.5)\)
• Mark down the **x-intercepts**: \( (0, 0) \) and \( (-3, 0) \)
• Note the **y-intercept**: which coincides at \( (0, 0) \)
• Remember that the parabola opens upwards due to \( a = 2 \), a positive value.

Using these points, draw a smooth symmetrical curve through them, creating a "U" shape. This visualization aids in understanding the function's behavior across different values of \( x \), showcasing its turning point at the vertex and demonstrating its direction of opening.