Problem 40 - Contemporary Precalculus

Question

In Exercises \(35-42,\) sketch the graph of the equation and label the vertex. $$y=x^{2}+8 x+6$$

Step-by-Step Solution

Answer

Answer: The vertex of the parabola is (-4, -10).

Step 1: Rewrite the equation in vertex form

To rewrite the quadratic equation in its vertex form, we need to complete the square. This will give us the equation in the form of $$y=a(x-h)^2+k$$, where (h, k) is the vertex. Here's the given equation: $$y=x^2+8x+6$$ Complete the square for the equation: $$y=(x^2+8x)+6$$ Take half of the coefficient of x, square it, and add and subtract it inside the parenthesis: $$y=(x^2+8x+16-16)+6$$ Factor the trinomial inside the parenthesis and simplify: $$y=(x+4)^2-10$$ So, the vertex form of the given equation is: $$y=(x+4)^2-10$$

Step 2: Find the vertex of the parabola

Now that we have the equation in vertex form, the vertex (h, k) can be found easily. Comparing with the standard vertex form $$y=a(x-h)^2+k$$, we have: $$h=-4$$ $$k=-10$$ So, the vertex of the parabola is: $$(-4, -10)$$

Step 3: Sketch the graph and label the vertex

Begin by plotting the vertex (-4, -10) on the coordinate plane. Since the parabola is in the form $$y=a(x-h)^2+k$$ with a positive coefficient for the x^2 term (in our case, a = 1), the parabola will open upwards. You can also find additional points on the graph by choosing arbitrary x-values and plugging them into the equation to find their corresponding y-values. For example, let's choose x = -6: $$y=(-6+4)^2-10$$ $$y=(2)^2-10$$ $$y=4-10$$ $$y=-6$$ So one point on the graph is (-6, -6). Similarly, find more points, plot them, and then draw the parabola through all the points, making sure it opens upwards and passes through the vertex. Finally, label the vertex (-4, -10) on your sketch. The completed graph will have a parabola opening upwards with the vertex labeled at the point (-4, -10).

Key Concepts

Vertex Form

The vertex form of a quadratic equation is a powerful way to understand the structure of a parabola. It is expressed as \( y = a(x-h)^2 + k \). Here, \((h, k)\) represents the vertex of the parabola. The vertex is a significant point because it's the minimum or maximum point of the parabola, depending on whether it opens upwards or downwards. In our exercise, the vertex form of the given equation \( y = x^2 + 8x + 6 \) was found to be \( y = (x+4)^2 - 10 \). This tells us that the vertex of the parabola is at \((-4, -10)\). This form makes it easy to plot the vertex directly onto a coordinate plane, giving us clear insight into both the orientation and position of the parabola.
To convert a quadratic equation from standard to vertex form, we often use a method called "completing the square." Understanding this method is crucial as it not only reveals the vertex but also centers the equation around that vertex, simplifying analysis and graphing.

Completing the Square

Completing the square is a technique used to transform a quadratic equation into vertex form. This method makes it simpler to identify the vertex of a parabola. The process involves creating a perfect square trinomial from a quadratic expression. Here's how it works:

• Start with the quadratic equation in standard form: \( y = x^2 + 8x + 6 \).
• Focus on the quadratic and linear terms: \( x^2 + 8x \).
• Take half of the linear coefficient (8), square it, giving 16, and add and subtract this number within the equation: \( y = (x^2 + 8x + 16 - 16) + 6 \).
• Rewrite the trinomial as a square: \( y = (x+4)^2 - 16 + 6 \).
• Simplify to get the vertex form: \( y = (x+4)^2 - 10 \).

By doing this, the quadratic is expressed neatly in vertex form \( y = a(x-h)^2 + k \), making it easier to draw and interpret the graph. The method "completing the square" essentially re-centers the quadratic equation around its vertex, which is particularly useful for analysis and graphing.

Parabola Graph

The graph of a quadratic equation forms a curve known as a parabola. The vertex of the parabola is a critical point that indicates its turning point in the graph. After completing the square, the equation \( y = (x+4)^2 - 10 \) tells us that the vertex at \((-4, -10)\) represents this turning point. Since the coefficient of the \( x^2 \) term is positive (\( a = 1 \)), the parabola opens upwards.

Plotting the parabola involves a few steps:

• Start by marking the vertex \((-4, -10)\) on the coordinate plane.
• Draw the axis of symmetry, which is a vertical line through the vertex, \( x = -4 \).
• Choose additional \( x \)-values (like \( x = -6\) ) to determine corresponding \( y \)-values and plot these points.
• Connect these points smoothly to depict the curve of the parabola. Make sure the curve is symmetrical around the axis of symmetry.

This graphical representation provides a visual understanding of how the equation behaves as the \( x \)-values change. Understanding the behavior of a parabola involves identifying its vertex, axis of symmetry, and the direction in which it opens. Together, these elements make graphing and analyzing quadratic equations more intuitive.