Problem 46 - Beginning and Intermediate Algebra

Question

Graph each equation using the vertex formula. Find the \(x\) - and \(y\) -intercepts. $$x=y^{2}+6 y-4$$

Step-by-Step Solution

Answer

The vertex of the parabola is (-3, -13), the x-intercept is at the point (-4, 0), and there are two y-intercepts: (0, 1) and (0, -7). The parabola opens horizontally.

Step 1: Rewrite the equation in vertex form

The given equation is \(x = y^{2} + 6y - 4\). To rewrite the equation in vertex form, we need to complete the square. The vertex form of a parabola is like: \[x = a(y-h)^2 + k\], where (h,k) is the vertex of the parabola. So first, write the equation as: \(x = y^2 + 6y + \square - 4 - \square\) To complete the square, we need to add and subtract the square of half of the coefficient of y in the equation, which is: \[\frac{6}{2} = 3\] So, the completed equation becomes: \(x = y^2 + 6y + (3)^2 - 4 - (3)^2\) Now, the equation becomes: \(x = (y^2 + 6y + 9) - 4 - 9\) \(x = (y + 3)^2 - 13\) Now, the equation is in vertex form.

Step 2: Find the vertex of the parabola

Now, the equation is in vertex form: \(x = (y + 3)^2 - 13\) From the vertex form equation, we can see that the vertex (h, k) of the parabola is (-3,-13).

Step 3: Find the x-intercepts

To find the x-intercepts, set y = 0 in the given equation: \(x = 0^2 + 6(0) - 4\) \(x = -4\) Thus, there is only one x-intercept, which is at the point (-4, 0).

Step 4: Find the y-intercepts

To find the y-intercepts, set x = 0 in the given equation: \(0 = y^2 + 6y - 4\) To solve this quadratic equation for y, we can use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Here, a = 1, b = 6, and c = -4. \(y = \frac{-6 \pm \sqrt{6^2 - 4(1)(-4)}}{2(1)}\) \(y = \frac{-6 \pm \sqrt{64}}{2}\) \(y = \frac{-6 \pm 8}{2}\) So, the two possible values for y are: \(y_1 = \frac{-6 + 8}{2} = \frac{2}{2} = 1\) \(y_2 = \frac{-6 - 8}{2} = \frac{-14}{2} = -7\) Thus, there are two y-intercepts: (0, 1) and (0, -7).

Step 5: Graph the parabola

Now that we have the vertex, x-intercept, and y-intercepts, we can graph the parabola. Plot the vertex (-3, -13), the x-intercept (-4, 0), and the y-intercepts (0, 1) and (0, -7). Since the equation is in the form \(x = a(y - h)^2 + k\), the parabola opens horizontally. Draw the parabola using the plotted points and converging on the vertex.

Key Concepts

Vertex Form

Understanding the vertex form of a parabola is crucial for graphing. In this context, vertex form makes it easier to identify the vertex of the parabola, which is the turning point. The formula is given as \(x = a(y-h)^2 + k\). Here, \((h,k)\) is the vertex. This form highlights the vertex’s position, enabling straightforward graph plotting.
For the given equation \(x = (y + 3)^2 - 13\), the vertex is \((-3, -13)\). This information tells you where the parabola turns, guiding the entire graph’s structure. Ensuring you have the equation in vertex form before graphing saves time and simplifies the process.

Completing the Square

Completing the square is a method used to convert a quadratic equation into vertex form. It makes graphing parabolas much easier by revealing their vertex directly.

To complete the square:

• Take the coefficient of \(y\), which is 6 in this equation.
• Find half of it (\(\frac{6}{2} = 3\)).
• Square the result (\(3^2 = 9\)).
• Add and subtract this square inside the equation to form a perfect square trinomial.

This manipulation transforms the equation from \(x = y^2 + 6y - 4\) into \(x = (y + 3)^2 - 13\). Completing the square offers a clear path to the vertex form, simplifying the visualization of the graph.

x-intercepts

The x-intercept(s) of a parabola occur where the graph crosses the x-axis. Here, you set \(y = 0\) in the equation to solve for \(x\).

For the equation \(x = y^2 + 6y - 4\):

• Substitute \(y = 0\), giving \(x = -4\).
• The x-intercept is thus at the point \((-4, 0)\).

This shows that the parabola crosses the x-axis at \((-4, 0)\). A single x-intercept indicates that the vertex is not on the x-axis, guiding how the graph curves.

y-intercepts

Finding y-intercepts involves setting \(x = 0\) in the equation and solving for \(y\). This tells you where the parabola intersects the y-axis.

Using the equation \(0 = y^2 + 6y - 4\):

• Apply the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• With \(a = 1\), \(b = 6\), and \(c = -4\), solve to get two values: \(y = 1\) and \(y = -7\).

These solutions give you the points \((0, 1)\) and \((0, -7)\) as the y-intercepts. This reveals two intersection points on the y-axis, providing precise targets for plotting the graph.