Sketching Of Quadratic Functions

Quadratic Functions is a very common but important chapter that is included in the IGCSE or IB syllabus. Relating a quadratic equation to its corresponding graph or sketching a quadratic function graph are very common questions that require a clear concept. I have seen many students having a phobia about sketching quadratic functions. Therefore, let us discuss this topic today. We will be taking four quadratic functions and will be seeing how to sketch the graphs of them. We will be discussing how to sketch a quadratic function that does not have any real root. We will also be seeing how to sketch quadratic functions when expressed in vertex form or the zero form.

1.

We have to sketch the graph of this quadratic function. The given quadratic function is in the general form, which is ax^2 + bx + c. The first thing that we find is the value of its discriminant. The finding of the discriminant is not essential for the sketching of any quadratic function. But I am adding this part to make you understand the relation between the value of the discriminant and the number of zeroes that a quadratic function will have.

If the discriminant, D is greater than 0, then the quadratic function has two unique real roots (the graph crosses the x-axis twice), else if D is equal to 0, the function has two identical real roots (the graph touches or meets the x-axis once), but if the value of D is less than 0, then the function does not have any real root. Therefore, if the value of D is less than 0, then the graph will either be entirely above the x-axis or else it will be completely below the x-axis. The finding of the discriminant is useful for another reason. You may have been given a function that does not contain any real root but since you have no idea of that, you keep on wasting your time in trying to factorise the given quadratic expression to get the roots.

The direction in which a parabola opens depends on the sign of 'a', the coefficient of x^2 in the given quadratic expression. If a > 0, the parabola opens upwards else if a < 0, the parabola opens downwards. For our function, the value of 'a' is 2 which is greater than 0, so the parabola opens upwards and is completely above the x-axis and does not have any real root.

Our next step will be to find the vertex of the parabola. The x-coordinate of the vertex of any parabola is given by the expression -b/(2a). The corresponding y-coordinate is found by substituting the numerical value of 'x' in the function expression.

So, the coordinates of the vertex of the parabola are (1.25, 0.875). We get the line of symmetry of the parabola from the x-coordinate of the vertex. Therefore, x = 1.25 is the required line of symmetry.

The constant term of the quadratic expression gives us the y-intercept. For our function, the constant term is 4. Therefore, (0, 4) is one of the points on the parabola.

For the other half of the parabola, we need a point on that half. Since the point (0, 4) is 1.25 units to the left of the line of symmetry, there should be a point that will be 1.25 units to the right of the line of symmetry, the y-value remaining the same. Therefore, its x-coordinate will be 1.25 + 1.25 = 2.5 and its y-coordinate will be equal to 4. So, we have got the coordinates of the point lying on the other half of the parabola. They are (2.5, 4). Now, we are in a position to complete the parabola.

2.

We may check, though not essential, whether this graph, parabola, will have any real root by finding the discriminant.

We find the value of the discriminant to be positive which means that there will be two unique real roots of the given quadratic function.

From the given expression we find the y-intercept to be equal to 10. That means, (0, 10) is a point on the parabola.

We have seen one method of finding the coordinates of the vertex. We will be learning the second method of finding the vertex using this function. We will be expressing the given quadratic function in vertex form. The vertex form of any quadratic expression is of the type a(x - h)^2 + k where (h, k) is the vertex. We will be converting the given general form of the quadratic expression to its vertex form using the completion of the square method.

Thus, we find the vertex of the parabola to be (2.5, -2.5). The quadratic function could have been given in the form of 2(x - 2.5)^2 - 2.5 as well. This is the equivalent vertex form of the given quadratic function. If the function is expressed in the vertex form, then to find the y-intercept, you have to expand the expression and find the constant term. I have started with the general form of the quadratic function because I wanted you to show the completion of the square method which is used while converting the general form to a vertex form of a quadratic function.

The line of symmetry is x = 2.5. Since there is a point with coordinates (0, 10) on the parabola, therefore by symmetry, there will be another point with coordinates (5, 10) on the parabola. so, the sketch of the given quadratic function is as follows --

3.

The given quadratic expression is in factorised form. We equate each factor to 0 and get the value of 'x' as a root of the given function. Therefore, the two roots of the given quadratic function are 2 and 4. So the coordinates of the corresponding points are (2, 0) and (4, 0). The line of symmetry will be exactly in the middle of the two x-values 2 and 4. Hence, the equation for the line of symmetry is x = 3. The vertex of the parabola will be lying on the line of symmetry, therefore, its x-coordinate will be equal to 3. The y-coordinate can be found by substituting the x value in the given function expression.

So, the coordinates of the vertex are (3, 1). The y-intercept can be obtained by substituting 0 for x in the function expression. We get the y-intercept as (4 - 0)(0 - 2) or -8. We have found four points lying on the parabola. Therefore, we are in a position to sketch the function.

We find the parabola is opening downwards. This could have been guessed by finding the sign of the term containing x^2 to be negative from the function expression.

It is important to remember that we have to sketch the quadratic functions, which means we have to show their shapes along with the coordinates of few important points. We could have factorised our previous function to get the zeroes along with the vertex and the y-intercept but that is not required for the sketch. So, we did not factorise the function.

4.

The leading negative sign suggests that the parabola will be opening downwards. This is the case of the repeated root. You may check the value of the discriminant. You will find it to be 0. Therefore, the parabola will be touching and not crossing the x-axis. The touching point will be the vertex of the parabola. The function expression is given in vertex form because the given expression can be written as

Therefore, the coordinates of the vertex are (1, 0). We can find the y-intercept by substituting 0 for x in the function expression. We get it to be (0, -1). Since x = 1 is the line of symmetry, therefore the third point of the parabola will be (2, -1). Now we sketch the graph.

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