Problem 16 - 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. $$ f(x)=2 x^{2}-4 $$

Step-by-Step Solution

Answer

The y-intercept is (0, -4). The axis of symmetry is x = 0, and the vertex is (0, -4).

Step 1: Find the y-intercept

The y-intercept of a function is the point where the graph crosses the y-axis. This happens when \( x = 0 \). Substitute \( x = 0 \) into the function to find \( f(0) \):\[f(0) = 2(0)^2 - 4 = -4\]So, the y-intercept is \( (0, -4) \).

Step 2: Determine the Axis of Symmetry and x-coordinate of the Vertex

The axis of symmetry for any quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). In this function, \( a = 2 \) and \( b = 0 \):\[x = -\frac{0}{2 \times 2} = 0\]Thus, the equation of the axis of symmetry is \( x = 0 \), which is also the x-coordinate of the vertex.

Step 3: Find the Vertex

To find the vertex, substitute the x-coordinate from the axis of symmetry into the original function:\[f(0) = 2(0)^2 - 4 = -4\]The vertex is \( (0, -4) \).

Step 4: Make a Table of Values Including the Vertex

Choose additional x-values around the vertex \( x = 0 \), and calculate corresponding y-values. For instance:\(x = -1 \Rightarrow f(-1) = 2(-1)^2 - 4 = -2\)\(x = 0 \Rightarrow f(0) = -4\)\(x = 1 \Rightarrow f(1) = 2(1)^2 - 4 = -2\)The table of values is:- \( (-1, -2) \)- \( (0, -4) \)- \( (1, -2) \).

Step 5: Plot the Graph

Using the table of values, plot the points \( (-1, -2), (0, -4), (1, -2) \) on a graph. Draw a symmetric parabola opening upwards with the vertex at \( (0, -4) \) and symmetry along the \( y \)-axis.

Key Concepts

y-intercept

The y-intercept is a crucial point for graphing any function. To find it, you set the x-value in the equation to zero and solve for y. For the quadratic function given, by substituting in zero for x in the expression \(f(x) = 2x^2 - 4\), you perform a straightforward calculation:

• First, substitute: \(f(0) = 2(0)^2 - 4\) which simplifies to \(-4\).

Hence, the y-intercept is the point \((0, -4)\). It represents where the parabola crosses the y-axis, and serves both as a starting guide for sketching the graph and as an anchor point for symmetry.
Understanding the y-intercept helps in verifying the graph's accuracy as this point should always align with the plotted opening of the parabola.

axis of symmetry

The axis of symmetry is an imaginary vertical line that passes through the vertex of the parabola and divides it into two mirror-image halves. For any quadratic function of the form \(ax^2 + bx + c\), this line can be determined using the formula \(x = -\frac{b}{2a}\). In the equation \(f(x) = 2x^2 - 4\), with \(b = 0\), the axis of symmetry simplifies significantly:

• Calculation: \(x = -\frac{0}{2 \times 2} = 0\).

Therefore, the axis of symmetry is \(x = 0\). This insight helps tremendously when graphing because it pinpoints exactly where the vertex will lie along the x-axis and allows for easier plotting because it sets a mirror line where points on either side align symmetrically.

vertex

The vertex is often referred to as the "turning point" of a parabola. It is the point at which the parabola changes direction—in this case, from descending to ascending. We've determined that the vertex lies on the axis of symmetry. To find it:

• Substitute the x-value from the axis of symmetry \(x = 0\) into the quadratic equation: \(f(0) = 2(0)^2 - 4\).
• The result is \(-4\), giving us the vertex at \((0, -4)\).

The vertex provides a vital reference point for graphing, as it indicates the lowest point of the parabola (since it opens upwards) and also represents the minimum value of the function. Recognizing the vertex's significance aids in drawing the parabola accurately, especially when combined with understanding the axis of symmetry.

table of values

Creating a table of values offers detailed insight into the shape and position of a parabola. It involves choosing x-values, computing the corresponding y-values, and plotting these coordinate pairs. With our equation \(f(x) = 2x^2 - 4\), you start with the vertex (which we know is at \((0, -4)\)) and select additional points around it to show the parabola's curve:

• For \(x = -1\), \(f(-1) = 2(-1)^2 - 4 = -2\), giving point \((-1, -2)\).
• For \(x = 1\), the calculation is similar, yielding point \((1, -2)\).

This table of values, when plotted, clearly shows the curvature and symmetry of the parabola. Providing these points ensures each part of the graph aligns correctly and visually confirms the properties determined algebraically, like symmetry and vertex position.

graphing parabolas

Graphing a parabola involves plotting points from the table of values and drawing a smooth curve through these points that reflects the parabola's properties. For the quadratic function at hand:

• Plot the known points: \((-1, -2)\), \((0, -4)\), and \((1, -2)\). These are points derived from substituting and solving x-values in the quadratic formula.
• Draw a parabolic curve that opens upwards, being certain it is symmetric concerning the y-axis (the identified axis of symmetry).

Using both the vertex and axis of symmetry as guides, check that each plotted parabola passes through calculated points, forming a "U" shape. Visual representation not only confirms mathematical predictions but also offers an intuitive understanding of how changes in the quadratic equation affect the parabola's orientation, width, and position.