Problem 45 - Beginning and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The vertex of the given equation is at (1, 3). The x-intercepts are \(x = 1 + \sqrt{3}\) and \(x = 1 - \sqrt{3}\). The y-intercept is \(y = 2\).

Step 1: Rewrite the equation

First, rewrite the given equation into the standard form of a quadratic equation: $$y = -x^2 + 2x + 2$$ Now, let's complete the square to rewrite the equation into the vertex form: $$y = -(x^2 - 2x) + 2$$ To complete the square, we need to add and subtract the square of half of the coefficient of x, which is \(\frac{(-2)²}{4} = 1\). So, the equation becomes: $$y = -(x^2 - 2x + 1 - 1) + 2$$ $$y = -((x - 1)^2 - 1) + 2$$ Now, rewrite the equation in the vertex form: $$y = -(x - 1)^2 + 3$$

Step 2: Identify the vertex

Now that we have the equation in the vertex form, we can easily identify the vertex. The vertex of the parabola is given by the coordinates (h, k), where: $$h = 1$$ $$k = 3$$ So the vertex of the parabola is (1, 3).

Step 3: Find the x-intercepts

To find the x-intercepts, we need to set y equal to 0 and solve for x: $$0 = -(x - 1)^2 + 3$$ Add \((x - 1)^2\) to both sides and subtract 3: $$(x - 1)^2 = 3$$ Now, take the square root of both sides: $$x - 1 = \pm\sqrt{3}$$ Add 1 to both sides: $$x = 1 \pm\sqrt{3}$$ So, the x-intercepts are \(x = 1 + \sqrt{3}\) and \(x = 1 - \sqrt{3}\).

Step 4: Find the y-intercepts

To find the y-intercepts, we need to set x equal to 0 and solve for y: $$y = -(0 - 1)^2 + 3$$ Simplify the equation: $$y = -(-1)^2 + 3$$ $$y = -(1) + 3$$ $$y = 2$$ Therefore, the y-intercept is \(y = 2\). #Conclusion#: The vertex of the given equation is at (1, 3). The x-intercepts are \(x = 1 + \sqrt{3}\) and \(x = 1 - \sqrt{3}\). The y-intercept is \(y = 2\).

Key Concepts

Vertex Form

The vertex form of a quadratic equation is particularly useful in graphing parabolas. It takes the form \[ y = a(x-h)^2 + k \] where \( (h, k) \) represents the vertex of the parabola.

• "a": This tells us which direction the parabola opens. If \(a\) is positive, the parabola opens upwards. If it's negative, it opens downwards.
• "h": This value, derived from completing the square, indicates the horizontal shift of the parabola from the origin.
• "k": This represents the vertical shift.

By putting a quadratic equation into vertex form, you can quickly determine the vertex, which is the highest or lowest point of the parabola, depending on its orientation.
In our exercise, the equation was converted into vertex form as:\[ y = -(x - 1)^2 + 3 \] The vertex is at the point \( (1, 3) \). This conversion not only helps in graphing but also simplifies the process of finding other features of the parabola.

x-intercepts

The x-intercepts of a parabola are the points where the graph crosses the x-axis. Finding these intercepts involves setting \(y = 0\) in the equation and solving for \(x\).
In our example problem, after transforming the equation into vertex form, we derived the equation:\[ 0 = -(x - 1)^2 + 3 \]Solving this equation involved:

• Moving all terms to one side: \((x - 1)^2 = 3\)
• Taking the square root of both sides, where \(x - 1 = \pm \sqrt{3}\)
• Adding 1 to isolate x: \(x = 1 \pm \sqrt{3}\)

This results in two x-intercepts: \(x = 1 + \sqrt{3}\) and \(x = 1 - \sqrt{3}\).
These values indicate where the parabola touches or crosses the x-axis. Knowing these values helps in sketching an accurate graph of the parabola, as they mark important points along the x-axis.

y-intercepts

The y-intercept is where the graph crosses the y-axis, and it can be found by setting \(x = 0\) in the quadratic equation and then solving for \(y\).
For our specific equation, substituting \(x = 0\) into the vertex form equation, we have:\[ y = -(0 - 1)^2 + 3 \]This simplifies to:

• \( y = -1 + 3 \)
• \( y = 2 \)

Thus, the y-intercept of the parabola is at \( y = 2 \).
This point tells us where the parabola crosses the y-axis, providing another key reference point when drawing the graph. Having clear y-intercepts aids in forming a complete picture of the parabola’s shape and behavior.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically in the form \[ ax^2 + bx + c = 0 \].They represent parabolic graphs, which can open upwards or downwards depending on the sign of the coefficient \(a\).

• Standard form: \( y = ax^2 + bx + c \)
• This form is very general and useful for identifying the coefficients and constant terms directly from the equation.
• Transforming into vertex form can provide insights into the graph's more detailed properties.

Quadratic equations appear in many areas of math and science. They model real-world situations like the path of a projectile, optimization problems, and areas of shapes. Understanding different forms of quadratic equations helps to solve them correctly whether dealing with equations symbolically, graphically, or numerically.