Problem 28 - Algebra A Combined Function

Question

Graph each quadratic function. Label the vertex and sketch and label the axis of svmmetrv. $$ G(x)=\frac{1}{5} x^{2} $$

Step-by-Step Solution

Answer

The vertex is at (0, 0); axis of symmetry is x = 0.

Step 1: Identify the Standard Form of Quadratic

The given function is \( G(x) = \frac{1}{5} x^2 \). This function is in the form \( G(x) = ax^2 + bx + c \), where \( a = \frac{1}{5} \), \( b = 0 \), and \( c = 0 \).

Step 2: Determine the Vertex

For a quadratic function in standard form \( ax^2 + bx + c \), the vertex \( (h, k) \) is found using \( h = -\frac{b}{2a} \) and \( k = G(h) \). Since \( b = 0 \), the vertex is at \( x = 0 \). Substituting \( x = 0 \) into \( G(x) \), we get \( G(0) = \frac{1}{5}(0)^2 = 0 \). Therefore, the vertex is at \( (0, 0) \).

Step 3: Find the Axis of Symmetry

The axis of symmetry for a quadratic function \( ax^2 + bx + c \) is always \( x = h \). Since the vertex is \( (0, 0) \), the axis of symmetry is \( x = 0 \).

Step 4: Sketch the Graph

Begin by plotting the vertex at \( (0, 0) \). Since \( a = \frac{1}{5} \) is positive, the parabola opens upwards. Select a few x-values such as \( x = 1, -1, 2, -2 \), calculate the corresponding \( y \) values, and plot these points. For \( x = 1 \), \( G(1) = \frac{1}{5}(1)^2 = \frac{1}{5} \). Similarly, plot \( (1, \frac{1}{5}) \), \( (-1, \frac{1}{5}) \), and other suitable points. Sketch the parabola through these points, reflecting across the axis of symmetry \( x = 0 \).

Step 5: Label the Graph Components

Clearly label the vertex \( (0, 0) \) on the graph. Also, draw and label the axis of symmetry line on the graph as \( x = 0 \). Ensure all plotted points are clear and the parabola is smooth.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a special point that can be considered the "tip" or the turning point of the parabola. For quadratic functions in the form \( ax^2 + bx + c \), determining the vertex is straightforward. The vertex can either be the lowest or highest point on the parabola, depending on whether it opens upwards or downwards.

To find the vertex, use the formula:

• h = \(-\frac{b}{2a}\)
• k = \( G(h) \)

In the exercise provided, we have \( b = 0 \) and \( a = \frac{1}{5} \). Plugging into the formula, \( h = 0 \), and substituting this back into \( G(x) \), we find \( k = 0 \). Thus, the vertex sits at the origin point \((0, 0)\).

The vertex is helpful because it tells us where the parabola turns and the axis of symmetry will pass through it.

Axis of Symmetry

The axis of symmetry in a quadratic graph is like an imaginary vertical line that divides the graph into two mirrored halves. It's a crucial feature because it indicates where the vertex is located in relation to the rest of the graph. The symmetry line ensures that each point on the parabola has a corresponding point directly across from it.

For any quadratic function \( ax^2 + bx + c \), the axis of symmetry can always be calculated as \( x = h \), where \( h \) is the x-coordinate of the vertex \((h, k)\). In our step-by-step solution, the vertex is \( (0, 0) \), so the axis of symmetry is the vertical line \( x = 0 \).

• This line is equally distant from both sides of the parabola.
• The graph looks the same on either side of it.

Remember, identifying the axis of symmetry is essential for graphing the parabola accurately, helping to place other points symmetrically around it.

Graphing Quadratic Equations

Graphing a quadratic equation involves understanding how to plot points that form a curve known as a parabola. This curve can either open upwards or downwards depending on the sign of the coefficient \( a \).

• If \( a \) is positive, the parabola opens upwards.
• If \( a \) is negative, the parabola opens downwards.

Starting with the vertex is a great first step when graphing. For our function \( G(x) = \frac{1}{5} x^2 \), the vertex is \( (0, 0) \), and since \( a = \frac{1}{5} \) is positive, the parabola opens upwards.

Next, choose points to plot. Select x-values on either side of the vertex, calculate y-values substituting each into the equation, and plot them. For example, for \( x = 1 \), \( G(1) = \frac{1}{5} \), leading you to plot the point \( (1, \frac{1}{5}) \). Do similarly for additional points like \( x = -1 \), \( x = 2 \), and reflect these points across the axis of symmetry \( x = 0 \).

Finally, connect the points smoothly to sketch the parabola. This process helps visualize the shape and direction of the quadratic function, making complex concepts more tangible.