Problem 3 - 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. $$ f(x)=-x^{2}+4 x-1 $$

Step-by-Step Solution

Answer

y-intercept: (0, -1); Axis of symmetry: x = 2; Vertex: (2, 3).

Step 1: Identifying the Quadratic Function

The given quadratic function is \( f(x) = -x^2 + 4x - 1 \). This is a standard quadratic equation of the form \( ax^2 + bx + c \), where \( a = -1 \), \( b = 4 \), and \( c = -1 \).

Step 2: Finding the y-intercept

To find the \( y \)-intercept, substitute \( x = 0 \) into the function.\[ f(0) = -(0)^2 + 4(0) - 1 = -1 \]. Therefore, the \( y \)-intercept is \((0, -1)\).

Step 3: Finding the Axis of Symmetry

The axis of symmetry for a quadratic equation \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Using the values \( a = -1 \) and \( b = 4 \), we have: \[ x = -\frac{4}{2(-1)} = 2 \]. So, the axis of symmetry is \( x = 2 \).

Step 4: Finding the x-coordinate of the Vertex

The \( x \)-coordinate of the vertex is the same as the axis of symmetry, which was found to be \( x = 2 \).

Step 5: Finding the y-coordinate of the Vertex

Substitute \( x = 2 \) back into the quadratic function to find the \( y \)-coordinate of the vertex: \[ f(2) = -(2)^2 + 4(2) - 1 = -4 + 8 - 1 = 3 \]. Therefore, the vertex is \((2, 3)\).

Step 6: Creating a Table of Values

Construct a table of values for \( x \) that includes the vertex. Choose \( x \)-values around the vertex, such as 0, 1, 2, 3, 4:- \( f(0) = -1 \)- \( f(1) = -(1)^2 + 4(1) - 1 = 2 \)- \( f(2) = 3 \) (vertex)- \( f(3) = -(3)^2 + 4(3) - 1 = 2 \)- \( f(4) = -(4)^2 + 4(4) - 1 = -1 \)

Step 7: Graphing the Quadratic Function

Use the \( y \)-intercept, vertex, and points from the table of values to graph the quadratic function. Plot the points: \((0, -1)\), \((1, 2)\), \((2, 3)\) (vertex), \((3, 2)\), \((4, -1)\). Draw the axis of symmetry at \( x = 2 \). Connect the points to form a parabola opening downwards because the coefficient of \( x^2 \) is negative.

Key Concepts

Understanding the Y-Intercept

The y-intercept of a quadratic function is the point where the graph crosses the y-axis. This is an important feature because it provides us with a starting value, where the input, or x-value, is zero. In mathematical terms, to find the y-intercept of a quadratic function, you simply substitute x=0 into the function and solve for y. For example, in the given function, \( f(x) = -x^2 + 4x - 1 \), when you plug in \( x = 0 \), it simplifies to \( f(0) = -1 \). This tells us that the y-intercept is at the point (0, -1).

• The y-intercept gives the initial value of the function on the y-axis.
• It is the point where x is zero.
• Always calculate the y-intercept for effective graphing.

Knowing the y-intercept helps in sketching out the graph more accurately.

Exploring the Axis of Symmetry

The axis of symmetry in a quadratic function is a vertical line that runs through the vertex of the parabola. This line splits the parabola into two mirror-image halves, making it a crucial guide when graphing the function. The formula to find the axis of symmetry in a standard quadratic equation \( ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Applying this formula to the given equation \( f(x) = -x^2 + 4x - 1 \), where \( a = -1 \) and \( b = 4 \), the axis of symmetry becomes \( x = 2 \).

• The axis of symmetry indicates the center line of the parabola.
• This line is crucial for graphing as it aligns with the vertex.
• Helps in finding the x-coordinate of the vertex directly.

The axis of symmetry is fundamental for understanding the symmetrical properties of the parabola.

Finding the Vertex of a Quadratic Function

The vertex of a quadratic function is the point at which the parabola changes direction. It is either the highest or lowest point of the curve, depending on whether the parabola opens upwards or downwards. For the given function, the vertex can be calculated using the axis of symmetry for the x-coordinate and substituting that back into the equation to find the y-coordinate. For \( f(x) = -x^2 + 4x - 1 \), the x-coordinate of the vertex is given by the axis of symmetry, \( x = 2 \). To find the y-coordinate, substitute x=2 back into the function: \[ f(2) = -(2)^2 + 4(2) - 1 = 3 \]Thus, the vertex is (2, 3).

• The vertex is the peak or valley of the parabola.
• The vertex provides critical information for plotting the graph.
• In this example, it indicates the maximum point since the parabola opens downwards.

Understanding the vertex is vital as it defines the most extreme point of the parabola.