Problem 22 - 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. $$ f(x)=\frac{1}{5} x^{2}+2 x+\frac{9}{5} $$

Step-by-Step Solution

Answer

Vertex: \((-5, -\frac{16}{5})\), opens upward; Intercepts: y-intercept \((0, \frac{9}{5})\), x-intercepts \((-9, 0)\) and \((-1, 0)\).

Step 1: Identify the Coefficients

For the quadratic function \( f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} \), we identify the coefficients as follows: \( a = \frac{1}{5} \), \( b = 2 \), and \( c = \frac{9}{5} \).

Step 2: Determine the Vertex

Use the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \). Substitute \( a \) and \( b \):\[ x = -\frac{2}{2 \times \frac{1}{5}} = -\frac{2}{\frac{2}{5}} = -5 \].To find the y-coordinate, substitute \( x = -5 \) back into the function: \[ f(-5) = \frac{1}{5}(-5)^2 + 2(-5) + \frac{9}{5} = 5 - 10 + \frac{9}{5} = -5 + \frac{9}{5} = -\frac{25}{5} + \frac{9}{5} = -\frac{16}{5} \].So, the vertex is \((-5, -\frac{16}{5})\).

Step 3: Determine Graph Orientation

The graph opens upward if \( a > 0 \) and downward if \( a < 0 \). Here, \( a = \frac{1}{5} > 0 \), so the graph opens upward.

Step 4: Find the Y-Intercept

The y-intercept is the point where the graph intersects the y-axis, which occurs at \( x = 0 \). So, plug in \( x = 0 \) into the equation:\[ f(0) = \frac{9}{5} \].Thus, the y-intercept is \( (0, \frac{9}{5}) \).

Step 5: Find the X-Intercepts

To find the x-intercepts, solve the equation \( \frac{1}{5}x^2 + 2x + \frac{9}{5} = 0 \). Multiply the entire equation by 5 to clear the fraction:\[ x^2 + 10x + 9 = 0 \].Factor the quadratic:\[ (x + 9)(x + 1) = 0 \].Set each factor to zero:\[ x + 9 = 0 \] or \[ x + 1 = 0 \].Thus, \( x = -9 \) or \( x = -1 \). So, the x-intercepts are \( (-9, 0) \) and \( (-1, 0) \).

Step 6: Graph the Function

Plot the vertex \((-5, -\frac{16}{5})\), the y-intercept \((0, \frac{9}{5})\), and the x-intercepts \((-9, 0)\) and \((-1, 0)\) on the coordinate plane. Since the parabola opens upwards, sketch the curve passing through these points.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots or x-intercepts of a quadratic equation. Quadratic equations have the general form \( ax^2 + bx + c = 0 \). The quadratic formula is written as:

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

This formula allows you to find the values of \( x \) where the parabola intersects the x-axis. It works for any quadratic equation, ensuring you can always find the solutions.
In practical use, knowing the coefficients \( a \), \( b \), and \( c \) is crucial. Substituting these values into the formula regales the roots. Additionally, checking the discriminant \( b^2 - 4ac \) can tell you about the solutions:

• If \( b^2 - 4ac > 0 \), there are two real and distinct roots.
• If \( b^2 - 4ac = 0 \), there is exactly one real root.
• If \( b^2 - 4ac < 0 \), the roots are complex and not real.

Parabola Graph

The graph of a quadratic function is a parabola. Depending on the coefficient \( a \), the parabola can either open upwards or downwards:

• If \( a > 0 \), the parabola opens upward, forming a 'U' shape.
• If \( a < 0 \), it opens downward, resembling an upside-down 'U'.

The vertex of the parabola is a critical point. It's the highest or lowest point of the graph, depending on the direction it opens. In quadratic functions, the vertex can be found using the formula for the x-coordinate \( x = -\frac{b}{2a} \), while the y-coordinate is obtained by substituting this x-value back into the equation.
Understanding the parabola's direction and vertex helps significantly in sketching its graph accurately.

X-Intercepts

X-intercepts are the points where the graph of a function crosses the x-axis. For the quadratic function, finding these points entails setting the function equal to zero and solving for \( x \).
For example, considering the function \( f(x) = \frac{1}{5}x^2 + 2x + \frac{9}{5} \), we transformed it to \( x^2 + 10x + 9 = 0 \) by multiplying through by 5 to eliminate fractions. Factor the resulting equation to find the solutions: \( (x + 9)(x + 1) = 0 \). Thus, the x-intercepts are at \( x = -9 \) and \( x = -1 \).
These intercepts are crucial for graphing, as they provide points through which the parabola passes. Placing these points on the graph helps illustrate the shape and position of the parabola accurately.

Y-Intercept

The y-intercept is where the graph touches the y-axis. For any function, it is obtained by setting \( x = 0 \) and solving for \( f(x) \). For the quadratic function given, substituting \( x = 0 \) results in \( f(0) = \frac{9}{5} \).
This point \((0, \frac{9}{5})\) is significant because it provides a clear starting spot for graphing the function. By plotting the y-intercept, along with the vertex and x-intercepts, we gain key points that help in sketching the complete parabola accurately.
Remember, the location of the y-intercept depends solely on the constant term \( c \) in the quadratic equation \( ax^2 + bx + c \). As such, it's easy to calculate but vital for accurate graphing.