Problem 72 - Contemporary Precalculus

Question

Find the points of intersection of the parabola \(4 x^{2}-8 x=2 y+5\) and the line \(y=15\)

Step-by-Step Solution

Answer

Answer: The points of intersection are \(\left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \(\left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\).

Step 1: Rewrite the equation of the parabola in terms of \(y\)

We are given the equation of the parabola as \(4 x^2 - 8 x = 2 y + 5\). We can rewrite this equation in terms of \(y\) by solving for \(y\): \(2 y = 4 x^2 - 8 x - 5 \Rightarrow y = 2 x^2 - 4 x - \dfrac{5}{2}\).

Step 2: Set the equations equal to each other

Now we will set the equation of the parabola equal to the equation of the line to find the \(x\)-values of the intersection points: \(2 x^2 - 4 x - \dfrac{5}{2} = 15\).

Step 3: Solve the quadratic equation

Now we need to solve the quadratic equation \(2 x^2 - 4 x - \dfrac{5}{2} - 15 = 0\). First, simplify the equation: \(2 x^2 - 4 x - 35 = 0\). Then, divide the equation by \(2\) to simplify it further: \(x^2 - 2x - 17.5 = 0\). Now, we can use the quadratic formula to solve for the \(x\)-values of the intersection points. The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our case, we have \(a = 1\), \(b = -2\), and \(c = -17.5\). Plugging in these values, we get: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-17.5)}}{2(1)}\). Simplifying further, we find: \(x = \dfrac{2 \pm \sqrt{72}}{2}\). So, the two possible \(x\)-values are: \(x_1 = \dfrac{2 + \sqrt{72}}{2}\) and \(x_2 = \dfrac{2 - \sqrt{72}}{2}\).

Step 4: Find the corresponding \(y\)-values

Now that we have the \(x\)-values of the intersection points, we can find the corresponding \(y\)-values by plugging them back into the equation of the line \(y = 15\). Thus, the coordinates of the intersection points are: \((x_1, y_1) = \left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \((x_2, y_2) = \left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\). So, the points of intersection of the parabola \(4 x^2 - 8 x = 2y + 5\) and the line \(y = 15\) are \(\left(\dfrac{2 + \sqrt{72}}{2}, 15\right)\) and \(\left(\dfrac{2 - \sqrt{72}}{2}, 15\right)\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation that involves terms up to the second degree (or power). They follow the general form: \[ax^2 + bx + c = 0\]where:\

• \(a\), \(b\), and \(c\) are constants,\
• \(x\) is the variable, and\
• \(a\) cannot be zero.\

In our problem, we had to solve the quadratic equation to find where the parabola intersects the line, leading to:\\[x^2 - 2x - 17.5 = 0\]
To find the solutions, we can utilize the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula gives us potential solutions (also known as the roots) for \(x\). These roots represent the x-values where the curve will intersect the line. Utilizing this method ensures that all potential intersections are considered.
The solutions are real and exist because the discriminant \( b^2 - 4ac \) is positive (72 in this case). When the discriminant is positive, two distinct solutions exist.

Parabola

A parabola is a symmetrical, U-shaped curve that is the graph of a quadratic equation. Parabolas can open either upward or downward, depending on the coefficient of the \(x^2\) term. If this coefficient is positive, the parabola opens upwards; if negative, downwards.
The standard form of a parabola can be represented as:\[y = ax^2 + bx + c\]
In our particular exercise, the equation of the parabola was:\[4x^2 - 8x = 2y + 5\]By rewriting and simplifying, we expressed it as:\[y = 2x^2 - 4x - \frac{5}{2}\]
This equation shows that the parabola opens upwards since the coefficient \(a = 2\) is positive. The vertex, or the highest or lowest point of the parabola, gives insight into its shape and is key to understanding its position relative to the x-axis and other curves like lines.

Line Equations

Lines are described by linear equations. Each point on the line satisfies these equations. They have the general form \[y = mx + c\]where

• \(m\) represents the slope of the line, and
• \(c\) is the y-intercept. This means the point where the line crosses the y-axis.

In this exercise, the line equation is straightforward:\[y = 15\]
This equation signifies a horizontal line cutting through the y-axis at 15. By comparing this line to the parabola, we can find points where they meet (intersect). To do this, we set the y-values of both equations equal. This finds the x-values where the parabola \(y = 2x^2 - 4x - \frac{5}{2}\) matches the line. Substituting these x-values back into the line equation confirms that their y-values are 15, providing the intersection points.