Problem 17 - Algebra 2

Question

Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=3 x^{2}+1 $$

Step-by-Step Solution

Answer

The y-intercept is 1, axis of symmetry is \(x = 0\), and the vertex is \((0, 1)\).

Step 1: Find the y-intercept

To find the \(y\)-intercept, we set \(x = 0\) in the equation of the function. This gives \(f(0) = 3(0)^2 + 1 = 1\). Therefore, the \(y\)-intercept is \(1\), so the point is \((0, 1)\).

Step 2: Determine the Axis of Symmetry

The axis of symmetry for a quadratic function \(f(x) = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). Here, \(a = 3\) and \(b = 0\), so the axis of symmetry is \(x = -\frac{0}{2 \times 3} = 0\). Thus, the equation of the axis of symmetry is \(x = 0\).

Step 3: Find the x-coordinate of the Vertex

The \(x\)-coordinate of the vertex is the same as the axis of symmetry, which is \(x = 0\).

Step 4: Make a Table of Values

To make a table of values, choose \(x\) values around the vertex (\(x = 0\)), calculate the corresponding \(y\) values, and include the vertex point.- \(x = -1\), \(f(-1) = 3(-1)^2 + 1 = 4\)- \(x = 0\), \(f(0) = 1\) (Vertex)- \(x = 1\), \(f(1) = 3(1)^2 + 1 = 4\)| \(x\) | \(f(x)\) ||---|---|| -1 | 4 || 0 | 1 || 1 | 4 |

Step 5: Graph the Function

Plot the points from the table of values: \((-1, 4)\), \((0, 1)\), and \((1, 4)\). Draw the parabola opening upwards from the vertex at \((0, 1)\). The axis of symmetry, \(x = 0\), shows the line about which the parabola is symmetric.

Key Concepts

y-intercept

The y-intercept of a quadratic function is found where the graph crosses the y-axis. At this point, the value of x is always 0. To find the y-intercept, simply substitute 0 for x in the function equation. For our function, \(f(x) = 3x^2 + 1\), substituting 0 gives:\[f(0) = 3(0)^2 + 1 = 1\]Therefore, the y-intercept is 1, and the corresponding point where the graph crosses the y-axis is at \((0, 1)\).

• The y-intercept reveals where the parabola touches the y-axis.
• Given by setting \(x = 0\), it directly reflects the constant term in the function equation.

axis of symmetry

The axis of symmetry in a quadratic function is a vertical line that divides the parabola into two mirror-image halves. This axis is critical because it passes through the vertex, offering a balance point for the parabola. For functions in the form \(f(x) = ax^2 + bx + c\), the axis of symmetry's equation is given by\[x = -\frac{b}{2a}\]In the example function \(f(x) = 3x^2 + 1\), the values are \(a = 3\) and \(b = 0\). Substituting these into the formula we get:\[x = -\frac{0}{2 \cdot 3} = 0\]

• The axis of symmetry is \(x = 0\), or the y-axis itself in this case.
• This line helps identify the vertex's x-coordinate, aiding in the graphing process.

vertex of a parabola

The vertex is the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. It represents the turning point of the graph.For a function in the standard form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the axis of symmetry:\[x = -\frac{b}{2a}\]With \(a = 3\) and \(b = 0\) for \(f(x) = 3x^2 + 1\), the x-coordinate of the vertex is 0. To find the y-coordinate, substitute \(x = 0\) back into the function:\[f(0) = 3(0)^2 + 1 = 1\]So, the vertex of the parabola is at the point \((0, 1)\), the lowest point in our upward-opening parabola.

• The vertex can serve as a reference for graphing the parabola.
• Knowing both the x and y coordinates is crucial for plotting accurately.

graphing quadratic functions

Graphing a quadratic function involves plotting points and drawing the shape of the parabola. Begin with points calculated from a table of values around the vertex.Using the function \(f(x) = 3x^2 + 1\), we have calculated:- For \(x = -1\), \(f(-1) = 4\)- For \(x = 0\) (vertex), \(f(0) = 1\)- For \(x = 1\), \(f(1) = 4\)Plot these points: - \((-1, 4)\)- \((0, 1)\) - \((1, 4)\)The plotted points should form a symmetrical shape with the axis of symmetry \(x = 0\).

• Connect the dots smoothly to form the parabola.
• The parabola opens upwards due to a positive \(a\) value in the function.
• The curve passes through the vertex and reflects symmetry around the axis.