Question

Graph each equation of the system. Then solve the system to find the points of intersection. $$ \left\\{\begin{array}{l} y=x^{2}+1 \\ y=4 x+1 \end{array}\right. $$

Short answer

The points of intersection are (0, 1) and (4, 17).

Step-by-step solution

Graph the Quadratic Equation

The first equation is given as y = x^2 + 1This represents a parabola opening upwards. To graph it, create a table of values for x and y. For example:| x | y || -2 | 5 || -1 | 2 || 0 | 1 || 1 | 2 || 2 | 5 |Plot these points on a coordinate plane and draw the parabolic curve.

Graph the Linear Equation

The second equation is given as y = 4x + 1This is a linear equation representing a straight line. To graph it, find the y-intercept and slope. The y-intercept is 1 and the slope is 4, meaning the line rises 4 units for every 1 unit it runs to the right. Plot the y-intercept at (0, 1) and use the slope to find a second point, such as (1, 5). Draw the line through these points.

Solve the System Algebraically

Set the equations equal to each other to find the points of intersection:x^2 + 1 = 4x + 1Subtract 1 from both sides:x^2 = 4xRearrange to set the equation to zero:x^2 - 4x = 0Factor out the common term x:x(x - 4) = 0Set each factor equal to zero:x = 0 or x = 4

Find the Corresponding y-Values

Substitute x = 0 and x = 4 back into either original equation to find the corresponding y-values. Using y = 4x + 1:For x = 0: y = 4(0) + 1 = 1For x = 4: y = 4(4) + 1 = 17Thus, the points of intersection are (0, 1) and (4, 17).

Key concepts

Graphing Quadratic Equations

A quadratic equation is a second-degree polynomial and is typically in the form of \(y = ax^{2} + bx + c\). The key feature of a quadratic equation is that it graphs to a parabola, which can either open upwards or downwards depending on the sign of the coefficient \a\. To graph a quadratic equation like \(y = x^{2} + 1\), follow these steps:

• Create a table of values. Select a few values for \x\ and calculate the corresponding \y\ values.
• For example, use \x\-values: -2, -1, 0, 1, 2. You will get y-values: 5, 2, 1, 2, 5 respectively.
• Plot these points (\(-2,5\), \(-1,2\), \(0,1\), \(1,2\), \(2,5\)) on the coordinate plane.
• Draw a smooth curve connecting these points to form a parabola.
  

Remember, the vertex of a parabola is its turning point, and it can be a high or low point depending on the direction of the parabola.

Solving Systems of Equations

A system of equations involves finding the values of variables that satisfy all equations in the system simultaneously. There are several methods to solve these systems:

• Graphical Method: Graph each equation and identify the points where the graphs intersect.
• Substitution Method: Solve one equation for one variable and substitute this expression into the other equation.
• Elimination Method: Combine the equations in a way that eliminates one of the variables, making it easier to solve for the remaining variable.
  

In our exercise, we used a combination of graphical and algebraic methods. Firstly, we graphed the equations \(y = x^{2} + 1\) and \(y = 4x + 1\). Then, we set the equations equal to find the intersections algebraically. Specifically, setting \(x^{2} + 1 = 4x + 1\) and solving for \x\ yielded the solutions \x = 0\ and \x = 4\. We substituted these values back into one of the equations to find the corresponding \y\ values and determined the intersection points.

Intersection of Graphs

The intersection of graphs refers to the points where the graphs meet or cross each other. These points represent the solutions to the system of equations. To find the intersection points:

• Graph both equations on the same coordinate plane.
• Identify the points where the graphs intersect visually.
• Confirm these points analytically by setting the equations equal and solving for the variables.

In our example, we graphed the quadratic equation \(y = x^{2} + 1\) and the linear equation \(y = 4x + 1\). By visual inspection, we saw they intersect at two points. To find these points precisely, we set \(x^{2} + 1 = 4x + 1\). Simplifying this resulted in \(x^{2} - 4x = 0\), which factors to \(x(x - 4) = 0\). Solving gives \x = 0\ and \x = 4\. Plugging these back into \(y = 4x + 1\), we obtained the intersection points as \ (0,1)\ and \ (4,17)\.
Understanding intersections helps solve systems and grasp relations between different equations.