Question

31-38, plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs, $$ \begin{array}{l} y=-x+1 \\ y=(x+1)^{2} \end{array} $$

Short answer

The graphs intersect at points (0, 1) and (-3, 4).

Step-by-step solution

Understand the Equations

We have two equations: \(y = -x + 1\) and \(y = (x + 1)^2\). The first equation is a linear equation, and the second is a quadratic equation. This will influence the shape of the graphs we plot.

Plot the Linear Equation

For the first equation \(y = -x + 1\), identify key points to plot on the graph: when \(x = 0, y = 1\) and when \(y = 0, x = 1\). Use these two points to draw the line for the equation.

Plot the Quadratic Equation

For the second equation \(y = (x + 1)^2\), identify the vertex, which is at \(x = -1\) and \(y = 0\). This is a parabola opening upwards. Plot a few additional points for accuracy: when \(x = 0, y = 1\) and when \(x = 1, y = 4\).

Find the Points of Intersection Algebraically

Set the equations equal to each other: \(-x + 1 = (x + 1)^2\). Rearrange it into a quadratic equation: \((x+1)^2 + x - 1 = 0\). Expand and simplify to get \(x^2 + 2x + 1 + x - 1 = 0\), leading to \(x^2 + 3x = 0\). Factor the equation as \(x(x+3) = 0\), giving solutions \(x = 0\) and \(x = -3\).

Calculate Corresponding \(y\)-Values for Intersections

For \(x = 0\), substitute into \(y = -x + 1\) to find \(y = 1\). For \(x = -3\), substitute into \(y = -x + 1\) to find \(y = 4\).

Confirm Points of Intersection

Check both points \((0, 1)\) and \((-3, 4)\) in both equations. For \(x = 0\), \(y = 1\) in both equations. For \(x = -3\), \(y = 4\) in both equations, confirming these are the points of intersection.

Key concepts

Linear Equation

Linear equations are the backbone of algebra, involving variables that are of the first degree or power of one. They form straight lines when graphed on a coordinate plane.
Given the equation \( y = -x + 1 \), it's a classic example of a linear equation.

• Slope-Intercept Form: Linear equations are often written in the form \( y = mx + b \).
• Here, \( m \) represents the slope and \( b \) the y-intercept.

The slope \( m = -1 \) implies the line falls as you move from left to right.
The y-intercept \( b = 1 \) means the line crosses the y-axis at 1.
Identifying these two key features helps in plotting the line accurately.

Quadratic Equation

Quadratic equations, unlike linear ones, involve terms up to the second degree. This means their highest variable power is two.
The equation \( y = (x+1)^2 \) is a quadratic equation.

• Standard Form: In general, quadratics are expressed as \( ax^2 + bx + c \).
• They form a parabolic curve when graphed.

In this case, the parabola opens upwards, and its vertex is located at \( (x = -1, y = 0) \).
Recognizing the vertex is crucial as it represents the peak or the lowest point of the parabola, aiding in accurate plotting.

Graph Plotting

Graph plotting involves placing points on a coordinate plane based on equations and joining them to visualize the equation's nature.
For the linear equation \( y = -x + 1 \), choose points like \( (0, 1) \) and \( (1, 0) \) to draw the line.

• For the quadratic equation \( y = (x+1)^2 \), start at its vertex \( (-1, 0) \).
• Plot other points like \( (0, 1) \) and \( (1, 4) \).

These points are used to outline the shape of the parabola. Observing the different shapes of lines and curves on the graph gives insights into how these equations behave in space.

Coordinate Plane

A coordinate plane is a two-dimensional surface on which we can graph equations. It consists of a horizontal axis (x-axis) and a vertical axis (y-axis).
Each point on the plane is represented as \( (x, y) \), where \( x \) represents horizontal placement and \( y \) vertical placement.
Understanding the coordinate plane is key to solving and graphing equations.

• Linear equations will present as straight lines across the plane.
• Quadratic equations form parabolas, which are curves that either open upwards or downwards depending on the coefficient's sign.

The points of intersection between different graphs on a coordinate plane can be used to find solutions to the equations.

Algebraic Solution

Algebraic solutions involve mathematical manipulation to find where the graphs intersect. This process is pivotal for finding the intersection of functions.
To find where the linear \( y = -x + 1 \) and quadratic \( y = (x+1)^2 \) intersect, set the equations equal to each other: \( -x + 1 = (x+1)^2 \).

• Simplify this to form \( x^2 + 3x = 0 \).
• By factoring, we obtain \( x(x+3) = 0 \), leading to solutions \( x = 0 \) and \( x = -3 \).

Substitute \( x \) values back into one of the original equations to obtain corresponding \( y \) values, which reveals the intersection points \((0, 1)\) and \((-3, 4)\).
These calculations show the steps of arriving at points where the graphs meet on the coordinate plane.