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Math Lesson 15.1.6 - Finding the Equation of a Parabola from its Graph

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Welcome to our Math lesson on Finding the Equation of a Parabola from its Graph, this is the sixth lesson of our suite of math lessons covering the topic of Quadratic Graphs Part One, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

Finding the Equation of a Parabola from its Graph

So far, we have looked at the equation of a parabola given and the task was to plot its graph by finding first the five special points described earlier in this tutorial. Now, we will deal with the reverse situation, i.e. when the graph of a parabola is given and we want to find its equation. In simpler words, now we want to use the parabola graph to find the coefficients a and b and the constant c. Again, we consider the five special points to extract any useful information from them regarding the coefficients and the constant of the parabola. Let's explain this point through an example.

Example 5

Find the equation of the parabola shown by the graph below.

Math Tutorials: Quadratic Graphs Part One Example

Solution 5

We can easily identify all the five special points we have discussed in theory. It is not necessary to find all of them but we will find as many as needed. Thus, the coordinates of points A and B which represent the x-intercepts of the graph are A(1, 0) and B(2, 0). Therefore, the x-coordinate of the vertex V (given the segment midpoint formula) is

xV = xA + xB/2
= 1 + 2/2
= 3/2

From here, we can write

xV = -b/2a = 3/2
-b/a = 3
b = -3a

From the graph, we can also see that the y-intercept is at C(0, 2). Therefore, we have c = 2, as the y-coordinate of the y-intercept is equal to the constant c. So far we have only partially identified the equation of the parabola as the coefficient a still unknown. We have

y = ax2 - 3ax + 2

We can find this coefficient by substituting the coordinates of any of the point identified so far (for example of point A) in the equation above. Thus, for x = 1 and y = 0 we obtain

0 = a ∙ 12 - 3a ∙ 1 + 2
0 = a - 3a + 2
0 = -2a + 2
2a = 2
a = 1

Therefore, the equation of the parabola shown in the figure is

y = 1 ∙ x2 - 3 ∙ 1 ∙ x + 2

or

y = x2 - 3x + 2

You have reached the end of Math lesson 15.1.6 Finding the Equation of a Parabola from its Graph. There are 6 lessons in this physics tutorial covering Quadratic Graphs Part One, you can access all the lessons from this tutorial below.

More Quadratic Graphs Part One Lessons and Learning Resources

Types of Graphs Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
15.1Quadratic Graphs Part One
Lesson IDMath Lesson TitleLessonVideo
Lesson
15.1.1Recalling Quadratic Equations
15.1.2Plotting Quadratic Graphs
15.1.3Is there an easy way to plot a Quadratic Graph?
15.1.4What if the Discriminant is Zero or Negative?
15.1.5What does the Sign of the Coefficient "a" Indicate for the Parabola?
15.1.6Finding the Equation of a Parabola from its Graph

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  3. Types of Graphs Revision Notes: Quadratic Graphs Part One. Print the notes so you can revise the key points covered in the math tutorial for Quadratic Graphs Part One
  4. Types of Graphs Practice Questions: Quadratic Graphs Part One. Test and improve your knowledge of Quadratic Graphs Part One with example questins and answers
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  6. Continuing learning types of graphs - read our next math tutorial: Quadratic Graphs Part Two

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