Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
9. Quadratic Systems
Continue to next subchapter

Exercise 13 Page 262

Find the vertex and the axis of symmetry of the parabola.

Graph:

Solutions: Approximately and

Practice makes perfect

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 grid. 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.
We found that the coordinate of the vertex is We will use the coordinate of the vertex to find its coordinate by substituting it into the given equation.
Simplify right-hand side
The coordinate of the vertex is so the vertex is at 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 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

,

Simplify right-hand side

Add and subtract terms

Equation (II) produced a true statement. In Equation (I), the answer is an approximation. This is because we could not state an exact answer just by looking at the graph, but we obtained a decent approximation. Therefore, we can say that is an approximated answer. Let's continue by checking

,

Simplify right-hand side

Subtract terms

Equation (II) produced a true statement. Just like before, in Equation (I) the answer is an approximation. This is because we could not state an exact answer just by looking at the graph, but we obtained a decent approximation. Therefore, we can say that is also an approximated answer.