Table of Contents

Last modified on March 28th, 2023

We can graph any quadratic equation using a graphing tool. Still, we need to make the graph manually to understand the logic behind its making.

Let us consider the simplest quadratic function f(x) = x2. Graphing this on a coordinate plane will give us the shape shown below.

Let us consider the most straightforward quadratic function f(x) = x^{2}. Graphing this on a coordinate plane will give us the shape shown below.

- When the value of ‘a’ is positive, the curve is U-shaped facing upwards
- When the value of ‘a’ is negative, the curve is inverted U-shaped, facing downwards
- When the value of ‘a’ increases, the curve squeezes inwards
- When the value of ‘a’ decreases, the curve bulge outwards

We prefer to graph quadratic equations when it is given in vertex form f(x) = a(x – h)^{2} + k, where (h, k) is the vertex and h = ${h=\dfrac{-b}{2a}}$, k = f(h). However, the quadratic equation is often represented in the standard form ax^{2} + bx + c = 0, which first needs to be converted to the vertex using the steps shown here.

The most important reason we use the vertex form is that the vertex, the lowest or the highest form of the graph, can be easily obtained. Also, since the graph is symmetrical about the axis x = h, it makes it easy to plot the graph.

Thus, from the graph, the ‘h’ shows us how far left or right the curve has shifted from x = 0. Similarly, ‘k’ shows how far up or down the curve has been shifted from y = 0.

Let us graph the quadratic function f(x) = x^{2} + 10x + 16 = 0. Comparing the given function with the quadratic function in the standard form f(x) = ax^{2} + bx + c, we get

a = 1, b = 10, c = 16

From the above values, we can say:

The graph will be U-shaped facing upwards

As the coefficient of x is 2, which is small, the graph will be a little squeezed than the x^{2} graph

Let us first calculate the vertex (h, k)

h = ${h=\dfrac{-b}{2a}}$ = ${\dfrac{-10}{2\times 1}}$= -5

k = f(-5) = (-5)^{2} + 10 × (-5) + 16 = -9

Thus, the vertex is at (-5, -9). We can now plot the graph manually.

It is the reverse way of what we have seen above. We can also solve quadratic equations given the graph corresponding to it.

Suppose we need to write the quadratic equation corresponding to the given graph.

From the graph, we can find the vertex as (-5, -16)

Substituting the value of the vertex in the vertex form of the quadratic equation, we get

f(x) = a(x – h)^{2} + k, here h = -5, k = -16

= a(x + 5)^{2} – 16

As the given parabola is also found to pass through the point (-1, 0) so

f(-1) = 0 …(1)

Again,

a(x + 5)^{2} – 16 at x = -1 is f(-1) = a(-1 + 5)^{2} – 16 …(2)

Equating (1) and (2)

a(-1 + 5)^{2} – 16 = 0

=> 16a – 16 = 15

=> 16a = 16

=> a = 1

Thus the quadratic equation is y = 1(x + 5)^{2} – 16

Last modified on March 28th, 2023