Problem 25 - Introductory Algebra for College Students

Question

Graph the parabola whose equation is given $$y=x^{2}+8 x+7$$

Step-by-Step Solution

Answer

The graph of the parabola \(y=x^{2}+8 x+7\) has a vertex at (-4,-9), the y-intercept at (0,7) and opens upwards.

Step 1: Determine the Vertex

The vertex of a parabola \(y = ax^2 + bx + c\) is given by the formula \((-\frac{b}{2a}, f(-\frac{b}{2a})\), where \(f(x)\) is the quadratic function. In this case, the vertex is \((-\frac{8}{2 * 1}, (1*(-\frac{8}{2*1})^2 + 8*-\frac{8}{2*1} + 7)\) = (-4,-9)

Step 2: Determine the y-intercept

The y-intercept is found by setting x = 0 in the equation. From the given equation \(y = x^2 + 8x + 7\), the y-intercept when x = 0 is (0,7)

Step 3: Plot Additional Points if necessary

You may need to plot additional points to get a more accurate parabola shape. Choose points to the left and right of the vertex and calculate their corresponding y values by substituting the x values into the equation.

Step 4: Draw the Parabola

After plotting the vertex, the y-intercept and any additional points as necessary, use these points to sketch an approximation of the parabola. The parabola should open upward since the coefficient of \(x^2\) (which is a) is positive, indicating it is a upward facing parabola.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a critical point that provides valuable information about its shape and position on the graph. To find the vertex of a quadratic equation in the form of \(y = ax^2 + bx + c\), we use the formula \(x = -\frac{b}{2a}\) to determine the x-coordinate of the vertex. Once we have the x-coordinate, we can substitute it back into the quadratic equation to find the y-coordinate. This gives us the full vertex point \((-\frac{b}{2a}, f(-\frac{b}{2a}))\). In our specific example \(y = x^2 + 8x + 7\), the vertex is calculated as follows:

• The x-coordinate is found using \(-\frac{8}{2 \times 1} = -4\).
• Substitute \(-4\) back into the equation: \(y = (-4)^2 + 8(-4) + 7 = -9\).

Therefore, the vertex of the parabola is \((-4, -9)\). This point is the minimum point of the parabola because the parabola opens upwards.

Y-intercept of a Parabola

Finding the y-intercept of a parabola allows us to understand where the graph crosses the y-axis. This is done by setting \(x = 0\) in the quadratic equation and solving for \(y\). Since this involves substituting \(x = 0\), it often simplifies the calculations.In our equation \(y = x^2 + 8x + 7\), let's find the y-intercept:

• Substitute \(x = 0\): \(y = 0^2 + 8(0) + 7 = 7\).

Thus, the y-intercept of this parabola is the point \((0, 7)\). This point tells us that the parabola crosses the y-axis at \(y = 7\). Identifying the y-intercept provides a starting point for graphing the parabola.

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, and its standard form is \(y = ax^2 + bx + c\). The graph of a quadratic equation is a curve called a parabola. Each parabola is symmetric around a vertical line that passes through its vertex. The term "quadratic" comes from "quad," meaning square, because the variable is squared \(x^2\).Key features of quadratic equations include:

• The coefficient \(a\) determines the direction the parabola opens (upward if \(a > 0\), downward if \(a < 0\)).
• It influences the "width" of the parabola; larger \(|a|\) values produce narrower parabolas.\(b\) and \(c\) affect the position and shape, but not the direction of opening.

In our example, the equation \(y = x^2 + 8x + 7\) has \(a = 1\), so the parabola opens upwards and is relatively standard in width.

Graphing a Parabola

Graphing a parabola involves plotting points and shaping the curve based on the quadratic equation you are working with. Start by plotting key features like the vertex and y-intercept. These points offer a framework for the parabola's shape.

• The vertex is \((-4, -9)\), which we found earlier.
• The y-intercept is \((0, 7)\). These critical points help anchor the parabola.

Next, plot additional points to ensure an accurate graph. Choose x-values near the vertex, substitute them into the equation, and calculate the corresponding y-values.The symmetry of the parabola around the vertex helps in reflecting these points across the axis of symmetry, improving accuracy.Finally, sketch the curve, ensuring it smoothly connects all the plotted points. The parabola should curve upwards since we previously noted that \(a = 1\). This process helps create a more precise and recognizable graph of your quadratic equation.