Problem 66 - College Algebra

Question

Use a graphing utility to graph the parabolas.Write the given equation as a quadratic equation in \(y\) and use the quadratic formula to solve for \(y .\) Enter each of the equations to produce the complete graph. $$y^{2}+10 y-x+25=0$$

Step-by-Step Solution

Answer

The equation becomes \(y = 5 \pm \sqrt{x}\). Draw the parabolic graph based on these two potential solutions.

Step 1: Rearrange the equation in form \(y^{2} = f(x)\)

Rearrange the given equation \(y^{2}+10y-x+25=0\) to form \(y^{2} = f(x)\). Here, we find \(f(x) = x-25\). So the equation becomes \(y^{2} = x-25-10y\).

Step 2: Use the quadratic formula to solve for \(y\)

Next we use the quadratic formula to solve for \(y\) in the equation \(y^{2} = x-25-10y\). The quadratic formula is given by \(y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b =-10\), and \(c =-x+25\). This yields, \(y=\frac{-(-10)\pm\sqrt{(-10)^{2}-(4)(1)(-x+25)}}{2(1)}\). This simplifies to \(y=\frac{10\pm\sqrt{100+4x-100}}{2}=\frac{10\pm\sqrt{4x}}{2}\). Thus, the roots are given by \(y=\frac{10\pm2\sqrt{x}}{2} = 5 \pm\sqrt{x}\).

Step 3: Graph the solutions

Plotting both potential solutions, \(y = 5 + \sqrt{x}\) and \(y = 5 - \sqrt{x}\) would produce the complete graph of the original equation. This will be a parabolic shape, opening to the right since we're squaring the x-term. While graphing, pay special care to the fact that the graph only exists for \(x \geq 0\), since \(x\) is under the square root.

Key Concepts

Parabolas

A parabola is a U-shaped curve that can open up, down, left, or right depending on its equation. When we discuss parabolas in the context of quadratic equations, we typically deal with equations in the form of

• vertical parabolas: \[ y = ax^2 + bx + c \]
• horizontal parabolas: \[ x = ay^2 + by + c \]

The vertex of the parabola is a crucial feature; it's the point where the parabola changes direction. For our equation \[ y^2 + 10y - x + 25 = 0 \] when rewritten, exhibits a horizontal parabola opening to the right. This is because the squared term is in \(y\), not \(x\). Understanding these directions and vertex points can help visualize and interpret the graph easily.

Graphing Utility

A graphing utility, such as a graphing calculator or an online graphing tool, is essential in visualizing complicated equations like parabolas. When using a graphing utility:

• Input your equation in its simplest form. For our purpose, this means rewriting equations as \[ y = 5 \pm \sqrt{x} \]
• Ensure all allowed values for \( x \) are considered. In our scenario, because \( \sqrt{x} \) means the square root of \( x \), graph only for \( x \geq 0 \)

The utility will show two potential parabolic graphs: one for each solution of y. This handy tool supports checking work quickly, especially when manually solving such equations can lead to errors or oversight in domain restrictions.

Completing the Square

Completing the square is a method used to solve quadratic equations, and it helps in converting the equation into a format that can be easily graphed or put into the quadratic formula. This involves making the equation reflect a perfect square trinomial. Let's walk through a brief overview:

• Start from a quadratic equation, for instance, like our \[ y^2 + 10y - x + 25 = 0 \]
• Focus on the \(y\) terms: \[ y^2 + 10y \]
• Add and subtract the square of half the linear coefficient to balance the equation. Here, this means adding and subtracting \( (10/2)^2 = 25 \)

This procedure transforms the equation into a "completed square," which allows us to handle it more easily with the quadratic formula. Understanding this technique is critical when a more structured graph or solution is required in algebra.

Quadratic Equation

Quadratic equations are fundamental algebraic expressions of the form \[ ax^2 + bx + c = 0 \].They result in parabolas when graphed. Solutions to these equations can be found through methods like factoring, completing the square, or using the quadratic formula. The quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] solves any quadratic equation, providing solutions, or 'roots.' Applied to our context:

• Given \[ y^2 = x - 25 - 10y \]
• \( a = 1 \), \( b = -10 \), and \( c = -x + 25 \)

Plug these into the formula to solve for two possible \(y\) values. These values are critical in preparing the graphs via solutions \( y = 5 + \sqrt{x} \) and \( y = 5 - \sqrt{x} \). Mastery of applying the quadratic formula is vital since it is a straightforward way to handle a wide range of problems in algebra.