Problem 14 - 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. $$ f(x)=x^{2}+2 x-3 $$

Step-by-Step Solution

Answer

The vertex is (-1, -4), the parabola opens upward, x-intercepts are (1, 0) and (-3, 0), y-intercept is (0, -3).

Step 1: Identify the quadratic function parameters

The standard form of a quadratic function is given by \( f(x) = ax^2 + bx + c \). For the function \( f(x) = x^2 + 2x - 3 \), identify \( a = 1 \), \( b = 2 \), and \( c = -3 \).

Step 2: Determine the direction the parabola opens

The sign of \( a \) determines whether the parabola opens upward or downward. Since \( a = 1 > 0 \), the parabola opens upward.

Step 3: Find the vertex using the vertex formula

The vertex \((h, k)\) of a parabola can be found using the formula \( h = -\frac{b}{2a} \) and \( k = f(h) \). Substitute \( a = 1 \) and \( b = 2 \) into the formula: \( h = -\frac{2}{2 \times 1} = -1 \). Calculate \( k = f(-1) = (-1)^2 + 2(-1) - 3 = -4 \). Therefore, the vertex is \((-1, -4)\).

Step 4: Calculate x-intercepts

X-intercepts occur where \( f(x) = 0 \). Solve \( x^2 + 2x - 3 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = 2 \), \( c = -3 \).\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2} \]Thus, \( x = 1 \) or \( x = -3 \). The x-intercepts are \( (1, 0) \) and \( (-3, 0) \).

Step 5: Calculate the y-intercept

The y-intercept is where \( x = 0 \). Calculate \( f(0) = 0^2 + 2(0) - 3 = -3 \). Thus, the y-intercept is \( (0, -3) \).

Step 6: Graphing the function

Using the vertex \((-1, -4)\), the x-intercepts \((1, 0)\) and \((-3, 0)\), and the y-intercept \((0, -3)\), plot these points on a coordinate plane. Draw a smooth, upward-opening parabola through these points to visualize the graph of the function \( f(x) = x^2 + 2x - 3 \).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point that represents the peak or the lowest point of the graph, depending on its orientation. For quadratic functions in the form \( f(x) = ax^2 + bx + c \), identifying the vertex helps in understanding the graph's shape and position. This is done using the vertex formula.
The vertex is given by the coordinates \((h, k)\), where:

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

In the quadratic function \( f(x) = x^2 + 2x - 3 \), substitute \( a = 1 \) and \( b = 2 \) into the formula to find \( h = -\frac{2}{2} = -1 \). Then, calculate \( k \) by evaluating \( f(-1) \), giving \( k = -4 \). Therefore, the vertex is at \((-1, -4)\). This point tells us the parabola reaches its lowest point here, giving the graph a minimum.

X-intercepts

X-intercepts are the points where the graph crosses the x-axis. These occur where \( f(x) = 0 \). Finding them involves solving the quadratic equation, which can often be done using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
For the equation \( x^2 + 2x - 3 = 0 \), use the values \( a = 1 \), \( b = 2 \), and \( c = -3 \) in the formula:

• Calculate the discriminant: \( b^2 - 4ac = 4 + 12 = 16 \)
• Solve for \( x \): \( x = \frac{-2 \pm 4}{2} \)

This computation results in \( x = 1 \) and \( x = -3 \), meaning the parabola meets the x-axis at \((1, 0)\) and \((-3, 0)\). These intercepts give important points where the graph truly intersects the axis and help trace the parabola's path.

Y-intercept

The y-intercept of a function is where the graph crosses the y-axis. It's the point where \( x = 0 \). This is easier to compute compared to x-intercepts because you simply substitute \( x = 0 \) into the function.
For the quadratic function \( f(x) = x^2 + 2x - 3 \), evaluate \( f(0) \):

• Substitute: \( 0^2 + 2 \cdot 0 - 3 = -3 \)

Hence, the y-intercept is \((0, -3)\). This place on the graph shows where the curve passes through the y-axis. It serves as a reference point when sketching the graph, as it's a fixed point easy to find by inspection.

Graphing Quadratic Equations

Graphing a quadratic equation like \( f(x) = x^2 + 2x - 3 \) involves plotting key points and connecting them smoothly. The critical points include the vertex, x-intercepts, and y-intercept. Here’s a step-by-step approach:
1. **Plot the Vertex**: Begin by plotting the vertex \((-1, -4)\) on the graph. This point is the turning or lowest point, where the direction of the parabola changes.

• The parabola opens upward because \( a = 1 \) is positive.

2. **Plot X-intercepts**: Mark \((1, 0)\) and \((-3, 0)\) on the x-axis. These points are where the graph intersects the x-axis.
3. **Plot the Y-intercept**: Add \((0, -3)\) to the plot. This is where the graph crosses the y-axis and gives further direction.
4. **Draw the Parabola**: Connect these points smoothly to form a U-shaped curve that opens upward. Ensure that the curve passes through all plotted points, giving a visual representation of the quadratic function.
By following these steps, you achieve a complete and accurate graph of the quadratic function, helping visual learners see the spatial relationships.