Problem 50 - Beginning and Intermediate Algebra

Question

Graph each equation using the vertex formula. Find the \(x\) - and \(y\) -intercepts. $$x=2 y^{2}+4 y+8$$

Step-by-Step Solution

Answer

In the given equation \(x=2y^2+4y+8\), we find the vertex using the vertex formula, obtaining the vertex at \((4, -1)\). The x-intercept is at the point \((8, 0)\), and there are no y-intercepts as the graph does not intersect the \(y\)-axis. The graph of the sideways parabola has the vertex \((4, -1)\), x-intercept at \((8, 0)\), and no y-intercepts.

Step 1: Rewrite the given equation in the form of a sideways parabola

A sideways parabola is in the form \(x = ay^2 + by + c\). Our given equation is: $$x=2y^2+4y+8$$ So, \(a = 2, b = 4, c = 8\).

Step 2: Find the vertex of the parabola

For a sideways parabola of the form \(x = ay^2 + by + c\), the vertex (x-vertex, y-vertex) is given by: \(y\)-vertex = \(-\frac{b}{2a}\) \(x\)-vertex = \(c - \frac{b^2}{4a}\) Here, \(a = 2, b = 4\), and \(c = 8\). Substituting these values, we get: \(y\)-vertex = \(-\frac{4}{2(2)} = -1\) \(x\)-vertex = \(8 - \frac{4^2}{4(2)} = 4\) So, the vertex of the parabola is at the point \((4, -1)\).

Step 3: Find the x-intercepts

To find the x-intercepts, we set \(y\) to 0 and solve for \(x\): $$x = 2(0)^2 + 4(0) + 8$$ $$x = 8$$ Here, the x-intercept is at the point \((8, 0)\).

Step 4: Find the y-intercepts

To find the y-intercepts, we set \(x\) to 0 and solve for \(y\): $$0 = 2y^2 + 4y + 8$$ Since there are no real solutions for \(y\) in this equation, the graph of the parabola does not intersect the \(y\)-axis, and there are no y-intercepts. Now, with all the required information, draw the graph of the given sideways parabola, with vertex \((4, -1)\), x-intercept at \((8, 0)\), and no y-intercepts.

Key Concepts

Vertex

In the world of parabolas, the vertex is a key element in understanding its shape and position. The vertex is the highest or lowest point on a parabola depending on its orientation. For a sideways parabola like in our equation, the vertex indicates the farthest left or right point.

To find the vertex in a sideways parabola of the form \(x = ay^2 + by + c\), we need to calculate both the x-vertex and the y-vertex.

• The y-vertex is provided by the formula \(-\frac{b}{2a}\), which gives the middle y-value.
• The x-vertex is given by the equation \(c - \frac{b^2}{4a}\), which figures out the farthest horizontal point.

In the given equation \(x = 2y^2 + 4y + 8\), substituting \(a = 2\), \(b = 4\), and \(c = 8\) into these formulas we find the vertex is located at \((4, -1)\). This point is where our sideways parabola changes direction, aligning perfectly according to the vertex formula.

X-Intercept

The x-intercept of a parabola is where the curve crosses the x-axis. At this point, the value of \(y\) is zero. It's like the curve's handshake with the x-axis.

To find the x-intercept for our equation, we set \(y = 0\) in the equation \(x = 2y^2 + 4y + 8\). This turns our equation into a straightforward calculation:

• Substituting \(y = 0\), we have \(x = 2(0)^2 + 4(0) + 8\).
• This simplifies to \(x = 8\).

Thus, the parabola crosses the x-axis at the point \((8, 0)\). This solitary intercept tells us a lot about the horizontal shift and positioning of the parabola on the coordinate plane.

Y-Intercept

In many contexts, a y-intercept is where the parabola crosses the y-axis. However, a sideways parabola like the one we are discussing may not necessarily have any y-intercepts. This happens when we cannot find a real \(y\) value where the curve meets the y-axis.

For our equation \(x = 2y^2 + 4y + 8\), identifying the y-intercepts requires setting \(x = 0\) and solving \(0 = 2y^2 + 4y + 8\).

• This quadratic equation in y has no real solutions.

Thus, our parabola does not intersect the y-axis, meaning there is no y-intercept. This information helps to visualize the shape of the parabola on the graph, recognizing it floats fully to one side of the y-axis.

Quadratic Equation

The quadratic equation is a foundational concept when discussing parabolas. In a standard form, a quadratic equation is expressed as \(ax^2 + bx + c = 0\). However, for our sideways parabola, it takes the form \(x = ay^2 + by + c\).

This tweak in variables indicates rotation in the parabola's orientation.
Understanding a quadratic equation involves recognizing its components:

• \(a\), \(b\), and \(c\) are constants dictating the parabola's openness, direction, and position.
• \(a\) influences whether the parabola opens to the left or right when it's alongside \(y^2\).

Each of these values dramatically shapes the graph:

• If \(a\) is positive, the parabola opens to the right.
• If \(a\) is negative, it opens to the left.

The equation \(x = 2y^2 + 4y + 8\) exemplifies a standard sideways parabola with its characteristic curves and vertex location. Mastering the quadratic equation forms the basis of not only recognizing but graphically representing such parabolas.