Math Lesson 15.1.3 - Is there an easy way to plot a Quadratic Graph?

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Welcome to our Math lesson on Is there an easy way to plot a Quadratic Graph?, this is the third 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.

Is there an easy way to plot a Quadratic Graph?

The best thing to do when trying to sketch a quadratic graph is to find some special points of the graph that allow you to visualise the width, direction and minimum/maximum position of the curve. We will know explain these special points and how to find them.

1. The x-intercept(s), if any. You can find these points by substituting y = 0 in the equation
   y = ax² + bx + c
   to obtain the corresponding quadratic equation with a single variable, this is a special case of the quadratic equation with two variables. This special equation has the general form
   ax² + bx + c = 0
   Then, if you are lucky and the discriminant is positive, you obtain two points of the graph, which correspond to the x-intercepts. They are A(x₁, 0) and B(x², 0), where x₁ and x² are the solutions of the above quadratic equation with one variable. These solutions (we also call them "roots") are calculated by the formulas
   x₁ = -b - √∆/2a and x₂ = -b + √∆/2a
   as described in the first paragraph of this tutorial. Therefore, since the coordinates of points A and B are A(x₁, 0) and B(x², 0) respectively, we obtain
   A(-b - √∆/2a,0) and B(-b + √∆/2a,0)
2. The vertex point V. Given that the graph is symmetrical when "folded" laterally according to the vertical line that passes through vertex V, we obtain two identical halves of the parabola. When doing this, the two x-intercepts also coincide. Therefore, we can use the half-segment formula explained in the previous tutorial to find the x-coordinate of vertex V. This is because the x-coordinate of the vertex coincides with the half-segment point of the segment AB. From the half-segment formula for the x-direction, we therefore obtain
   x_V = x_A + x_B/2
   Given that
   x_A = x₁ = -b - √∆/2a
   and
   x_B = x₂ = -b + √∆/2a
   we obtain
   x_V = 1/2 ∙ (-b - √∆/2a + -b + √∆/2a)
   = -b - √∆ - b + √∆/4a
   = -2b/4a
   = -b/2a
   Substituting this value obtained for x in the original equation
   y = ax² + bx + c
   yields for the vertical coordinate of the vertex V:
   y_V = a ∙ (-b/2a)² + b ∙ (-b/2a) + c
   = a ∙ b²/4a² - b²/2a + c
   = b²/4a - b²/2a + c
   = b²/4a - 2b²/4a + c
   = -b²/4a + c
   = -b²/4a + 4ac/4a
   = -b² + 4ac/4a
   = -(b² - 4ac)/4a
   = -∆/4a
   Therefore, the coordinates of the vertex V are
   V(-b/2a, -∆/4a)
3. Another point of the parabola that is easy to find is the y-intercept. Since the x-coordinate of any y-intercept is always zero, we can solve the original quadratic equation for x = 0. In this way, we obtain y = c. Therefore the coordinates of the y-intercept (say point C) are C(0, c).
4. Optional! You can also find the symmetrical point to the y-intercept to obtain a kind of symmetry in the points and therefore make the graph more accurate. For this, we again use the half-segment formula, where x_V is the half-segment point, x = 0 is one of the endpoints and the other endpoint has to be found. In this case, we need to find only the x-coordinate of this new point (say D), as the y-coordinate is the same as that of point C (because of the symmetry). This will be clearer after solving the example below.

Using these 5 points, you can plot a quadratic graph with a satisfactory level of accuracy. Look at the following example.

Example 1

Plot the graph of the quadratic equation

Solution 1

We have a = 1, b = -5 and c = 4. The discriminant Δ therefore is

Since the discriminant is positive, the equation

has two distinct roots, so the parabola produced by the original quadratic equation has two x-intercepts: x_A and x_B. They are:

and

The corresponding y-values for points A and B are zero, as they are x-intercepts. Therefore, we obtain the first two points of the graph:

Now, let's find the coordinates of the vertex V of the parabola. We have

and

Therefore, the third point of the graph is

Next, let's find the coordinates of the fourth point of the graph, i.e. of point C, which is the point where the parabola intercepts the y-axis. Thus, for x = 0, we have y = c. Given that c = 4, the coordinates of point C are: C(0, 4).

Last, let's find the coordinates of point D, which is symmetrical with point C in respect to the vertical line drawn from point V. Applying the formula of the segment midpoint (here the midpoint is x_V),

for the horizontal direction only, we obtain after the substitutions:

The corresponding y-value of point D therefore is

Hence, the fifth point of the graph is D(5, 4).

Now, let's insert all the above points into the coordinate system. We have

Connecting these points smoothly and continuing in that way in both directions yields the following graph:

You have reached the end of Math lesson 15.1.3 Is there an easy way to plot a Quadratic 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

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