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Quadratic graph Graphing quadratic functions How to factor quadratic equationsHere you will learn about solving quadratic equations graphically, including how to find the roots of a quadratic function from a graph, how to use this method to solve any quadratic equation by drawing a graph, and then how to solve a quadratic equation from a graph that is given.
Students first learn how to work with quadratic functions and their graphs in Algebra I and expand that knowledge as they work with polynomials in Algebra II.
Solving quadratic equations graphically is a strategy to find the roots of a quadratic equation by using its graph, which is a parabola. The roots of a quadratic function are the values of x that make the equation true and equal to 0. They are also called the zeros of the function.
Letβs find the roots of the quadratic equation, x^2-6 x+8=0 by using the graph.
Using the graph, notice that when x=2 and x=4 the y -value or f(x) is equal to 0.
Therefore, the roots of the function, x^2-6 x+8=0 are x=2 and x=4.
So, you can conclude that the x -intercepts are the roots of the quadratic equation.
Meaning that the roots are the zeros and are the x -intercepts.
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DOWNLOAD FREEYou can also use the graph of a quadratic function to solve an equation such as x^2=4
In order to do this, treat the left hand side and the right hand side as two separate functions to graph, similar to when solving a system of equations graphically.
In this case, since the left hand side is x^2, graph the equation, y=x^2 and since the right hand side is equal to 4, graph the equation, y=4.
Notice the points of intersection, they represent the solutions to the equation.
The points of intersection are, (- \, 2, \, 4) and (2, \, 4), so the solutions to the equation, x^2=4 are x=2 and x=- \, 2.
There are other strategies you can use to find the solution to the equation such as taking the square root or factoring.
How does this relate to high school math?
In order to find the solutions of a quadratic equation using a graph:
Find the solutions of the equation x^{2}-4=0 graphically.
No rearrangement is needed in this case.
2Graph the quadratic equation.
Write y=x^2-4 and draw the graph.
See also: Graphing quadratic functions
3Identify the \textbf{x} -intercept(s).
The quadratic equation intersects the x -axis at two points, x=- \, 2 and x=2. So, the quadratic has two x -intercepts.
4State the root(s) to the quadratic.
Since the x -intercepts are at x=- \, 2 and x=2, the roots or solutions to the quadratic equation are, x=- \, 2 and x=2.
Find the solutions of the equation x^{2}+6x+9=0 graphically.
Rearrange the equation so that one side \bf{= 0} (if necessary).
No rearrangement is needed in this case.
Graph the quadratic equation.
Write y=x^{2}+6x+9 and sketch the graph.
Identify the \textbf{x} -intercept(s).
The quadratic equation touches the x -axis at one point, x=- \, 3. So, the quadratic equation has one x -intercept.
State the root(s) to the quadratic.
Since there is only one x -intercept, there is only one real root or solution to the quadratic.
The root or solution to the equation x^{2}+6 x+9=0 is x=- \, 3.
Find the solutions of the equation x^{2}-2x+4=0 graphically.
Rearrange the equation so that one side \bf{= 0} (if necessary).
No rearrangement is needed in this case.
Graph the quadratic equation.
Write y=x^{2}-2x+4 and sketch the graph.
Notice how the parabola does not touch or intersect the x -axis. This means that the quadratic equation has no x -intercepts.
Identify the \textbf{x} -intercept(s).
The parabola does not intersect the x -axis, so there are no x -intercepts.
State the root(s) to the quadratic.
Since the quadratic equation, x^{2}-2x+4=0, has no x -intercepts, it does not have any real roots or real solutions.
Find the solutions of the equation 12+x=x^{2} graphically.
Rearrange the equation so that one side \bf{= 0} (if necessary).
In this case, rearrange the quadratic equation so that it is equal to 0. Subtract the x^2 from both sides of the equation.
\begin{aligned}& 12+x=x^2 \\\\
& - \, x^2+12+x=x^2-x^2 \end{aligned}
Then write the quadratic equation in standard form, a x^2+b x+c=0 .
- \, x^2+x+12=0
Graph the quadratic equation.
Notice that βaβ is negative so the parabola will be upside down when graphed.
The quadratic equation intersects the x -axis at two points, so there are two x -intercepts.
Identify the \textbf{x} -intercept(s).
The x -intercepts are x=- \, 3 and x=4.
State the root(s) to the quadratic.
Since there are two x -intercepts, there are two roots or solutions. The roots to the quadratic, 12+x=x^2, are x=- \, 3 and x=4.
Find the solutions of the equation x^2-12 x=- \, 11 graphically.
Rearrange the equation so that one side \bf{= 0} (if necessary).
In this case, add 11 to both sides of the equation.
x^2-12 x+11=- \, 11+11
Write the equation in standard form, a x^2+b x+c=0 .
x^2-12 x+11=0
Graph the quadratic equation.
The quadratic equation intersects the x -axis at two points.
Identify the \textbf{x} -intercept(s).
The x -intercepts are x=1 and x=11.
State the root(s) to the quadratic.
Since the quadratic equation has two x -intercepts, it has two real roots. The roots or solutions to the quadratic, x^2-12 x=- \, 11, are x=1 and x=11.
In order to find the solutions of systems of linear and quadratic equations graphically:
Find the solution(s) of the quadratic equation graphically, x^2=9.
Identify the quadratic equation and linear equation to be graphed.
In this case, one side of the equation represents the quadratic to graph and the other side of the equation represents the linear equation. So, the quadratic to graph is y=x^2 and the linear equation to graph is the horizontal line y=9.
Graph both the quadratic equation and the linear equation.
Identify the points of intersection.
y=x^2 and y=9 intersect at two points.
State the solution(s).
The solution(s) are the x values of the points of intersection. So, in this case, the solutions are x=3 and =- \, 3.
Find the solutions of the equation x^{2}-2x+4=2x+4 graphically.
Identify the quadratic equation and linear equation to be graphed.
In this case, one side of the equation represents the quadratic to graph and the other side of the equation represents the linear equation. So, the quadratic to graph is y=x^2-2 x+4 and the linear equation to graph is y=2 x+4.
Graph both the quadratic equation and the linear equation.
The quadratic equation and the linear equation intersect at two points.
Identify the points of intersection.
State the solution(s).
Since there are two points of intersection, there are two solutions to the equation.
The solutions are the x -values of the point of intersection.
(0, \, 4) and (4, \, 12).
So, the solutions are x=0 and x=4.
1. Find the solutions of the equation x^{2}-9=0 graphically.
The quadratic equation is already set equal to 0. Graph the quadratic equation and identify the x -intercepts.
The x -intercepts are x=- \, 3 and x=3. So, since the quadratic has two x -intercepts it has two real roots (or solutions). The solutions are x=- \, 3 and x=3.
2. Use the graph of the quadratic equation x^{2}-5x+4=0 to determine the solutions.
From the graph, you can see that the parabola intersects the x -axis at two points, when x=- \, 1 and x=- \, 4. Since the quadratic has two x -intercepts, it has two solutions. The solutions or roots are the x -intercepts, x=- \, 1 and x=- \, 4.
3. Use the graph of the quadratic to determine the solutions of the function, y=x^2-2 x-15.
Looking at the graph of the quadratic function, y=x^2-2 x-15, you can see that the parabola intersects the x -axis at two points.
This means that the quadratic has two x -intercepts, at x=- \, 3 and x=5. The x -intercepts are the roots or solutions to the quadratic equation which means that the solutions are x=- \, 3 and x=5.
4. Which graph demonstrates the solutions to the quadratic equation, y=x^2-8 x+16.
Set the quadratic equation equal to 0.
\begin{aligned}& y=x^2-8 x+16 \\\\ & 0=x^2-8 x+16 \end{aligned}
Then use a strategy to graph the quadratic and identify the x -intercept(s).
In this case, there is only 1 \, x -intercept because the parabola touches the x -axis at one point.
The point where the parabola touches the x -axis is at x=4. Since the x -intercept is x=4, the root or solution to the quadratic equation is also x=4.
5. Use the graph to determine the solution(s) of the equation, 2=x^2+4 x-3.
x=5 and x=1
x=- \, 5 and x=1
x=- \, 5 and x=- \, 1
Using the given equation, 2=x^2+4 x-3, graph the left hand side and the right hand side. In this case, the left hand side of the equation is 2 so graph y=2 and the right hand side of the equation is x^2+4 x-3 so graph y=x^2+4 x-3.
The solution(s) are the x -coordinates of the points of intersection.
The solutions are x=- \, 5 and 1.
6. Use the graph to determine the solution(s) of the equation, x^2-9=3 x-9.
x=0 and x=3
x=- \, 9 and x=0
Using the given equation, x^2-9=3 x-9, graph the left hand side and the right hand side. In this case, the left hand side of the equation is x^2-9 so graph y=x^2-9 and the right hand side of the equation is 3x-9 so graph y=3x-9.
The solution(s) are the x -coordinates of the points of intersection.
You can sketch the quadratic equations by hand using one of the strategies learned in algebra 1 such as making a table of values, using the vertex and axis of symmetry or using the vertex form of a quadratic equation. You can also use a graphing calculator or digital graphing calculator to sketch the quadratic equation.
It depends on what the actual graph of the function looks like. If it clearly intersects the x -axis and the values by which the graph intersects the x -axis are easy to determine, then using the graph is a good strategy to use. However, if the graph of the function does not intersect the x -axis or it intersects the x -axis not at a whole number, then using a different strategy would be more effective.
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