Question

plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs. $$ \begin{array}{l} y=-2 x+3 \\ y=3 x^{2}-3 x+12 \end{array} $$

Short answer

The graphs of the equations do not intersect at any real points.

Step-by-step solution

Understand the Functions

First, identify the type of equations you have. The first equation, \(y = -2x + 3\), is a linear equation, producing a straight line. The second equation, \(y = 3x^2 - 3x + 12\), is a quadratic equation, forming a parabola.

Plot the Linear Equation

On a coordinate plane, start by plotting the linear equation \(y = -2x + 3\). You can find two points easily. For example, when \(x = 0\), \(y = 3\); and when \(x = 1\), \(y = 1\). Draw a straight line through these points to represent the equation.

Plot the Quadratic Equation

Plot the quadratic equation \(y = 3x^2 - 3x + 12\). Find a few points for this equation. For example, when \(x = 0\), \(y = 12\); when \(x = 1\), \(y = 12\); and when \(x = 2\), \(y = 18\). Connect these points smoothly to form a parabola.

Determine Points of Intersection

To find the points of intersection, set the two equations equal to each other: \(-2x + 3 = 3x^2 - 3x + 12\). Simplify this to form \(3x^2 - x + 9 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=3\), \(b=-1\), \(c=9\) to solve for \(x\).

Solve Using the Quadratic Formula

Plug the values into the quadratic formula: \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(9)}}{2(3)}\), which becomes \(x = \frac{1 \pm \sqrt{1 - 108}}{6}\). Calculate the discriminant to find that it is negative, \(1 - 108 = -107\), indicating no real solutions.

Verify Intersection Points

Since the discriminant is negative, this means the quadratic and the linear function do not intersect at any real points. Hence, there are no points of intersection in the real number plane.

Key concepts

Graphical Representation

When dealing with systems of equations, graphical representation is a powerful tool to visualize solutions. It involves plotting two or more equations on the same coordinate plane. Each equation's graph shows a set of solutions:

• A linear equation forms a straight line.
• A quadratic equation forms a curve known as a parabola.

Graphical methods allow you to see where these solutions intersect, which indicates the common solutions to the system.
In our example, we are tasked with plotting a linear and a quadratic equation together. The intersection points would typically appear as the coordinates where the line meets the curve. However, sometimes these graphs do not intersect on the real number plane, indicating no real solutions. This happens when the line and the parabola either are parallel or do not intersect due to their geometric arrangement.
Graphically, this can be determined by observing if they touch or cross each other anywhere on the graph.

Linear Equations

The first type of equation we encounter in this example is a linear equation:

• The general form is: \(y = mx + c\).

Here, \(m\) represents the slope of the line, and \(c\) is the y-intercept, indicating where the line crosses the y-axis.
The equation given, \(y = -2x + 3\), describes a straight line with a slope of -2 and a y-intercept at (0, 3).
To plot a linear line, select two points by choosing values for \(x\) and then computing the corresponding \(y\). In this case:

• When \(x = 0\), \(y = 3\).
• When \(x = 1\), \(y = 1\).

Once these points are plotted, draw a line through them. This line visually shows all possible solutions to the linear equation.

Quadratic Equations

The other equation in our system is quadratic, typically written as \(y = ax^2 + bx + c\).

• This form generates a parabola, a symmetrical curve with an axis of symmetry.
• The parabola opens upward if \(a > 0\) and downward if \(a < 0\).

In the example, the equation \(y = 3x^2 - 3x + 12\) opens upward due to \(a > 0\).
To graph this:

• Choose values for \(x\) and calculate \(y\).
• Plot these (\(x, y\)) points.

Using points like \((0, 12)\), \((1, 12)\), and \((2, 18)\), you can draw the parabola.
However, a critical understanding is finding intersection points with other graphs.
If such intersection cannot be found, as in the case here due to a negative discriminant, it implies that the graphs don't meet at any points on the real plane. This outcome is visual evidence of no shared solutions among the considered variables.