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Solving Linear-Quadratic Systems

Solving Linear-Quadratic Systems 1.5 - Solution

arrow_back Return to Solving Linear-Quadratic Systems

To solve the system of equations by graphing, we will draw the graph of the quadratic function and the linear function on the same coordinate system. Let's start with the parabola.

Graphing the Parabola

To graph the parabola, we first need to identify and For this equation we have that and Now, we can find the vertex using its formula. To do this, we will need to think of as a function of Let's find the -coordinate of the vertex.
Evaluate
We will use the coordinate of the vertex to find its coordinate by substituting it into the given equation.
The coordinate of the vertex is Therefore, the vertex is the point With this, we also know that the axis of symmetry of the parabola is the line Next, let's find two more points on the curve, one on each side of the axis of symmetry.

Both and are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.

Graphing the Line

Let's now graph the linear function on the same coordinate plane. For a linear equation written in slope-intercept form, we can identify its slope and -intercept The slope of the line is and the intercept is

Finding the Solutions

Finally, let's try to identify the coordinates of the points of intersection of the parabola and the line.

It looks like the points of intersection occur at and

Checking Our Answer

Checking the answer
To check our answers, we will substitute the values of the points of intersection in both equations of the system. If they produce true statements, our solution is correct. Let's start with
,
Identity Property of Addition
Equation (I) and Equation (II) both produced true statements. Therefore, is a correct solution. Let's continue by checking
,
Add terms
Equation (I) and Equation (II) produced true statements again. Therefore, is also a correct solution.