Problem 25 - Algebra A Combined Function

Question

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=2 x^{2}+4 x+5 $$

Step-by-Step Solution

Answer

Vertex: \((-1, 3)\); Opens upward; Y-intercept: \((0, 5)\); No X-intercepts.

Step 1: Identify the Form of the Quadratic Function

The given function is in the standard quadratic form: \( f(x) = ax^2 + bx + c \). Here, \( a = 2 \), \( b = 4 \), and \( c = 5 \).

Step 2: Determine the Direction of the Parabola

Since the leading coefficient \( a = 2 \) is positive, the parabola opens upward.

Step 3: Find the Vertex of the Parabola

The vertex of a parabola given in standard form \( y = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substitute \( b = 4 \) and \( a = 2 \) to find \( x \):\[x = -\frac{4}{2(2)} = -1\]Substitute \( x = -1 \) back into the function to find \( y \):\[f(-1) = 2(-1)^2 + 4(-1) + 5 = 2 - 4 + 5 = 3\]Thus, the vertex is \((-1, 3)\).

Step 4: Find the Intercepts

**Y-intercept**: Set \( x = 0 \) in \( f(x) = 2x^2 + 4x + 5 \).\[f(0) = 2(0)^2 + 4(0) + 5 = 5\]The y-intercept is \((0, 5)\).**X-intercepts**: Set \( f(x) = 0 \) and solve \( 2x^2 + 4x + 5 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[x = \frac{-4 \pm \sqrt{4^2 - 4(2)(5)}}{2(2)} = \frac{-4 \pm \sqrt{16 - 40}}{4} = \frac{-4 \pm \sqrt{-24}}{4}\]Since the discriminant \(-24\) is negative, there are no real x-intercepts.

Step 5: Sketch the Graph

The graph is an upward-opening parabola with the vertex at \((-1, 3)\) and the y-intercept at \((0, 5)\). There are no x-intercepts.

Key Concepts

Vertex of a Parabola

In a quadratic function of the form \( f(x) = ax^2 + bx + c \), the vertex is a key point that defines the graph's shape and position. It serves as either the highest or the lowest point, depending on the parabola's direction. The formula to find the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). For the given function \( f(x) = 2x^2 + 4x + 5 \), we substitute \( b = 4 \) and \( a = 2 \) into this formula, calculating \( x = -\frac{4}{2(2)} = -1 \).
To find the y-coordinate, substitute \( x = -1 \) back into the function:

• \( f(-1) = 2(-1)^2 + 4(-1) + 5 = 3 \)

Thus, the vertex of this function is located at the point \((-1, 3)\), which is the lowest point since the parabola opens upwards.

Parabola Direction

The direction of a parabola—whether it opens upwards or downwards—is determined by the sign of the leading coefficient \( a \) in the quadratic function \( f(x) = ax^2 + bx + c \). If \( a > 0 \), the parabola opens upward, resembling a U shape. Conversely, if \( a < 0 \), it opens downward, resembling an upside-down U. In our scenario, the function \( f(x) = 2x^2 + 4x + 5 \) has \( a = 2 \), which is positive. Thus, this parabola opens upward.
This upwards opening signifies that the vertex represents the function's minimum point. Understanding the parabola's direction is crucial for sketching its graph and predicting its behavior.

Quadratic Formula

The quadratic formula is a reliable tool for finding the roots or x-intercepts of a quadratic equation \( ax^2 + bx + c = 0 \). It is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the problem at hand, we use this formula to determine whether real x-intercepts exist for the function \( f(x) = 2x^2 + 4x + 5 \). Substituting \( a = 2 \), \( b = 4 \), and \( c = 5 \), we calculate:

• \( x = \frac{-4 \pm \sqrt{4^2 - 4(2)(5)}}{4} \)
• \( = \frac{-4 \pm \sqrt{16 - 40}}{4} \)
• \( = \frac{-4 \pm \sqrt{-24}}{4} \)

The term inside the square root, called the discriminant, is \(-24\). Since it is negative, this indicates there are no real x-intercepts, as real roots arise only when the discriminant is zero or positive.

Intercepts of Quadratic Functions

Intercepts provide valuable points on the graph of a quadratic function that intersect the axes. There are two types: x-intercepts and y-intercepts.

**Y-intercept**: This occurs where the graph crosses the y-axis, found by setting \( x = 0 \) in the function. For \( f(x) = 2x^2 + 4x + 5 \):

• \( f(0) = 5 \)

The y-intercept is at \((0, 5)\).

**X-intercepts**: These are the points where the graph crosses the x-axis, obtained by setting the function equal to zero. However, as we previously calculated, the discriminant is negative in this case, meaning there are no real x-intercepts.
Understanding intercepts aids in the sketching and analysis of the graph. The y-intercept is a concrete point, whereas the x-intercepts are absent here, guiding our understanding of the parabola's path.