Question

Use analytical methods to find the following points of intersection. Use a graphing utility to check your work.Find the point(s) of intersection of the parabola \(y=x^{2}+2\) and the line \(y=x+4\).

Short answer

Answer: The points of intersection between the parabola and the line are (2, 6) and (-1, 3).

Step-by-step solution

Set equal the expressions for y from both equations

The parabola is defined by the equation \(y=x^2+2\) and the line by the equation \(y=x+4\). To find the point of intersection, we need to find the values of \(x\) and \(y\) where both equations are equal. Since both equations are equal to \(y\), we can set the expressions on the right side of the equal sign equal to each other: $$x^2+2=x+4$$

Solve for x

In order to solve for \(x\), we need to rearrange the equation to have all terms on one side of the equal sign: $$x^2+2-x-4=0$$ $$x^2-x-2=0$$ Now, we must solve this quadratic equation for x. Since the equation doesn't seem to have any easy factors, we'll use the quadratic formula: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ In our equation, \(a=1\), \(b=-1\), and \(c=-2\). Plugging these values into the quadratic formula, we get: $$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-2)}}{2(1)}$$ $$x=\frac{1\pm\sqrt{1+8}}{2}$$ $$x=\frac{1\pm\sqrt{9}}{2}$$ $$x=\frac{1\pm 3}{2}$$ This gives us two solutions for \(x\): \(x=2\) and \(x=-1\).

Find corresponding y values and points

Now that we have the \(x\) values, we can find the corresponding \(y\) values by plugging the \(x\) values into either the parabola equation or the line equation (since they are equal at the points of intersection). We will use the parabola equation: For \(x=2\): $$y=2^2+2=4+2=6$$ So one point of intersection is \((2, 6)\). For \(x=-1\): $$y=(-1)^2+2=1+2=3$$ So the other point of intersection is \((-1, 3)\).

Check results with a graphing utility

Finally, we can check our results using a graphing utility to ensure that the intersections we found are correct. By graphing the parabola \(y=x^2+2\) and the line \(y=x+4\), we can visually confirm if the two points we found, \((2, 6)\) and \((-1, 3)\), are indeed the points of intersection. Upon checking with a graphing utility, it confirms that our answers are correct. Therefore, the points of intersections between the given parabola and the line are \((2, 6)\) and \((-1, 3)\).

Key concepts

Parabola

A parabola is a U-shaped curve that can open upwards or downwards in a two-dimensional graph. It is the graph of a quadratic function, which is typically expressed in the form \(y = ax^2 + bx + c\). Here, the coefficients \(a\), \(b\), and \(c\) determine the specific shape and position of the parabola:

• The coefficient \(a\) affects the direction and the width of the parabola.
• If \(a\) is positive, the parabola opens upwards, forming a "U" shape.
• If \(a\) is negative, it opens downwards, forming an upside-down "U".
• Larger values of \(a\) make the parabola "narrower," while smaller values make it "wider."

The parabola in our exercise is presented by the equation \(y = x^2 + 2\), which means \(a = 1\), \(b = 0\), and \(c = 2\). This particular parabola opens upwards and its vertex, the lowest point, is at \((0, 2)\).
When we intersect this with a line, we aim to find the exact points where the line crosses the parabola.

Linear Equation

A linear equation represents a straight line in a two-dimensional graph and it involves no exponents greater than one. It typically appears in the form \(y = mx + b\), where:

• \(m\) is the slope of the line, indicating how steep the line is.
• \(b\) is the y-intercept, showing the point where the line crosses the y-axis.

For the line \(y = x + 4\) in our exercise, the slope \(m = 1\), which means the line rises one unit for every unit it moves horizontally. The y-intercept is \(4\), meaning the line crosses the y-axis at the point \((0, 4)\).
Finding the intersection of this line with a parabola means calculating the \((x\), \(y\)) coordinates that satisfy both the quadratic and the linear equations simultaneously.

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which is any equation that can be re-arranged into the standard form \(ax^2 + bx + c = 0\). The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to find the values of \(x\) that make the quadratic equation equal to zero, effectively solving for \(x\) when factoring is not straightforward.
In the exercise, the quadratic equation \(x^2 - x - 2 = 0\) was solved using the quadratic formula:

• The coefficient \(a = 1\), \(b = -1\), and \(c = -2\) were identified from the equation.
• Using these coefficients in the formula, the discriminant (\(b^2 - 4ac\)) was calculated as \(9\), which is a perfect square, indicating two real and distinct solutions.
• Thus, the solutions for \(x\) were \(2\) and \(-1\).

These solutions give us the \(x\)-coordinates of the points where the parabola and the line intersect, which were then used to find the corresponding \(y\)-values.