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3. Quadratic Functions & Equations
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United States of AmericaTennessee
Algebra 2

3.03 Linear-quadratic systems

Lesson
Practice
Lesson

Concept summary

A system of equations is a set of equations that have the same variables.

The solution to a system of equations is any ordered pair that makes all of the equations in the system true. For graphs this will be the point(s) of intersection. Solutions can be found algebraically or graphically.

x
y
The line and parabola have no points of intersection so the system has no solution.
x
y
The line and parabola have one point of intersection so the system has one solution.
x
y
The line and parabola have two points of intersection so the system has two solutions.

The solution to a system of equations in a given context is viable if the solution makes sense in the context, and is non-viable if it does not make sense.

Worked examples

Example 1

Consider the following systems of equations:

\begin{cases} y= x^{2} - 2 x - 3 \\ y= - x + 3 \end{cases}

a

Graph the equations on the same coordinate plane.

Approach

The solution(s) to a system of equations can be represented graphically as their point(s) of intersection. We can use technology to graph the two equations, or, if drawing them by hand, it will be useful to first fill out a table of values for both equations.

x-3-2-101234
y=x^{2} - 2 x - 31250-3-4-305
x-3-2-101234
y=-x+36543210-1

Solution

-5
-4
-3
-2
-1
1
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3
4
5
x
-5
-4
-3
-2
-1
1
2
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5
y

Reflection

Notice that from the table of values both function have the columns with the points \left(-2,5\right) and \left(3,0\right). We want these points of intersection to be visible on our graph. We also want the vertex of the parabola, \left(1,-4\right), to be visible.

We want to ensure that the x-values on our graph cover at least the interval -3 \leq x \leq 4 and the y-values on our graph cover at least the interval -5 \leq y \leq 6.

b

Identify the coordinates of the solution(s) to the system of equations.

Solution

The points of intersection occur at \left(-2, -5\right) and \left(0, 3\right), so these coordinate pairs will be the solutions to the system of equations.

Reflection

We can check our work by solving the system of equations algebraically, by equating both equations and solving for x

\displaystyle x^2-2x-3 \displaystyle =\displaystyle -x+3Use that y=y, reflexive property
\displaystyle x^2-x-3\displaystyle =\displaystyle 3Add x to both sides
\displaystyle x^2-x-6\displaystyle =\displaystyle 0Subtract 3 from both sides
\displaystyle \left(x+2\right)\left(x-3\right)\displaystyle =\displaystyle 0Factor the quadratic

Using the zero product law we can see the two solutions to this new quadratic are x=-2 and x=3

We can now substitute these into one of the given equations to find the corresponding y-value.

Example 2

Find the solution(s) for the following linear-quadratic system of equations.\begin{cases} y = 3 x + 1 \\ y = x^{2} - 5x \end{cases}

Approach

We can approach this graphically or algebraically.

-4
-3
-2
-1
1
2
3
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5
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9
x
-5
5
10
15
20
25
y

But we can see from the graph that the points of intersection are not clearly identifiable. In cases like this, an algebraic approach is preferable. As both equations are already in terms of y we can use the substitution method to solve.

Solution

\displaystyle 3x+1 \displaystyle =\displaystyle x^2-5xUse that y=y, reflexive property
\displaystyle 1\displaystyle =\displaystyle x^2-8xSubtract 3x from both sides
\displaystyle 0\displaystyle =\displaystyle x^2-8x-1Subtract 1 from both sides

This equation is not easily factorable, and so we will use the quadratic formula to find the solutions.

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-\left(-8 \right) \pm \sqrt{\left(-8\right)^2-4\left(1 \right)\left(-1\right)}}{2\left( 1\right)}Substitute a=1, b=-8, c=-1
\displaystyle =\displaystyle \frac{8\pm \sqrt{64+4}}{2}Evaluate the square and products
\displaystyle =\displaystyle \frac{8\pm \sqrt{68}}{2}Evaluate the difference in the radicand
\displaystyle =\displaystyle \frac{8\pm 2\sqrt{17}}{2}Simplify the radical
\displaystyle =\displaystyle \frac{8}{2}\pm \frac{2\sqrt{17}}{2}Rewrite as two fractions
\displaystyle =\displaystyle 4\pm \sqrt{17}Simplify the quotients

We have found the x-coordinates of the points of intersection. We can substitute these into either equation to find the corresponding y-coordinate.

\displaystyle y\displaystyle =\displaystyle 3x+1
\displaystyle y\displaystyle =\displaystyle 3\left(4-\sqrt{17}\right)+1
\displaystyle y\displaystyle =\displaystyle 13+3\sqrt{17}

So one solution is \left(4-\sqrt{17}, 13-3\sqrt{17}\right), and using the same method we find the other solution is \left(4+\sqrt{17}, 13+3\sqrt{17}\right)

Example 3

Forrest and his child Gustavo are driving remote control cars. Forrest is practicing turning his car in a parabolic arc, while Gustavo races his car in a straight line.

They are driving on the local basketball court before anyone gets there. Consider one corner as the origin, the long side of the court as the x-axis and the short side of the court as the y-axis.

A first quadrant coordinate plane with no scales on its axes. A diagram of a basketball court is plotted as a rectangle using a part of the x axis as the length and y axis as the width. Two semi circles are drawn, one using a shorter segment in the middle of the y axis and another in the opposite side of the rectangle. A smaller circle is drawn at the center of the rectangle and a segment from the horizontal side to the x axis divides the circle in half.

Forrest's car follows the curve y=\left(x-4\right)^2+3 and Gustavo's car follows the line y=-\dfrac{3}{5}x+9, where x and y are in feet.

a

Graph the two paths on the same coordinate plane.

Approach

The equation for Forrest's car is in vertex form, so we can see the direction of opening from the coefficient, then plot the vertex, and substitute in another value for x to get the shape.

The equation for Gustavo's car is a linear function, so we can identify the y-intercept, find another point on the line using the slope, and then graph the line through those two points.

Solution

1
2
3
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5
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x
1
2
3
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5
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9
y
b

Use technology to determine if the cars' paths would cross and, if so, the coordinates of the points.

Approach

We can start by inputting the two equations for the cars' paths into a graphing calculator. We already have a sketch, so know we should end up with a parabola and a line.

A screenshot of the Geogebra Graphing Calculator with the functions y equals x minus 4 all squared plus 3, and y equals minus 3x over 5 plus 9. Talk to your teacher for more information.

We can estimate the coordinates by eye, but we should determine the coordinates more precisely using an intersection tool.

Solution

From the graph we can see that the graphs intersect twice, so there are two points of intersection. To find the coordinates of these points we can use the intersection tool. For the built-in GeoGebra graphing calculator, we need to click on the point of intersection.

A screenshot of the Geogebra Graphing Calculator with the one of the intersections of the functions highlighted. Talk to your teacher for more information.
A screenshot of the Geogebra Graphing Calculator with the other intersection of the functions highlighted. Talk to your teacher for more information.

The two points where the paths of the cars would cross are: \left(1.78, 7.93\right) and \left(5.62, 5.63\right).

Reflection

We can confirm this algebraically as well:

\displaystyle \left(x-4\right)^2+3\displaystyle =\displaystyle -\frac{3}{5}x+9Use that y=y, reflexive property
\displaystyle x^2-8x+16+3\displaystyle =\displaystyle -\frac{3}{5}x+9Square the binomial
\displaystyle x^2-8x+19\displaystyle =\displaystyle -\frac{3}{5}x+9Combine like terms
\displaystyle 5x^2-40x+95\displaystyle =\displaystyle 3x+45Multiply both sides by 5
\displaystyle 5x^2-43x+50\displaystyle =\displaystyle 0Move all terms to one side

The solutions of this quadratic equation would give us the x-coordinates for any points of intersection between the two graphs.

The discriminant of this equation is \left(-43\right)^2-4 \cdot 5 \cdot 50=849>0 which means there are two real solutions.

We could then use the quadratic formula to get the x-values of the points of intersection and then substitute into the original equations to get the y-coordinates.

Outcomes

A2.N.Q.A.1

Use units as a way to understand real-world problems.*

A2.A.REI.B.4

Solve a system consisting of a linear equation and a quadratic equation in two variables algebraically, graphically, and using technology.

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP5

Use appropriate tools strategically.

A2.MP6

Attend to precision.

A2.MP8

Look for and express regularity in repeated reasoning.

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