How does the discriminant affect the graph?

Hint: We will try to solve “Siddhartha Acharya Formula” for a quadratic equation and then we will try to interpret the behavior of discriminant of a quadratic equation.

Complete step-by-step solution:
So, We found the roots for the quadratic equation of \[a{x^2} + bx + c = 0\].
And, the roots of the equation are:
\[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\] or \[x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\].
So, it is pretty much clear that the nature of the roots depends on the value of the square root part which is known as “Discriminant”.
So, Discriminant is \[ = \sqrt {{b^2} - 4ac} \].
Now, if the value of \[({b^2} - 4ac)\] is greater than zero, then the value of the roots will be definitely the real numbers, whether it can be positive or negative.
So, in that case discriminant should have a positive value.
Now, if the value of \[({b^2} - 4ac)\] is equal to zero, then the value of the roots will have only one value, which states that the values of the roots will be equal to \[ - \dfrac{b}{{2a}}\].
So, the equal roots can have any real numbers.
Now, if the value of \[({b^2} - 4ac)\] is less than zero, then the value of the discriminant will become an imaginary number.
It means that the roots will be imaginary numbers.
So, the nature of the graph depends on the value of the discriminant.
So, we can draw the following chart by the value of discriminant:


Note: Nature of the roots always depends on the value of discriminant.
Discriminant tells us that whether the equation has one solution or, two solutions or, no solution.
Proof for descriminant:
Let's say \[a{x^2} + bx + c = 0\] is a quadratic equation, where \[a,b,c\] are constant and we have to solve the equation for \[x\].
So, we can write this equation as following:
\[a{x^2} + bx + c = 0\]
Now divide both the sides of the equation by \[a\], we get:
\[\dfrac{a}{a}{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\]
\[ \Rightarrow \] \[{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\].
Now we can rewrite the above equation in the following form:
\[ \Rightarrow {(x)^2} + 2 \times \dfrac{b}{{2a}} \times x + \dfrac{c}{a} = 0\].
Or, we can add a magnitude of zero by\[{\left( {\dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2}\]: \[ \Rightarrow {(x)^2} + 2 \times \dfrac{b}{{2a}} \times x + {\left( {\dfrac{b}{{2a}}} \right)^2} - {\left( {\dfrac{b}{{2a}}} \right)^2} + \dfrac{c}{a} = 0\]
Now, taking only constant terms into the R.H.S, we get:
 \[ \Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2}}}{{4{a^2}}} - \dfrac{c}{a}\]
Simplifying the constant terms, we get:
\[ \Rightarrow {\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}\]
Now, taking the square root on the both sides:
\[ \Rightarrow \left( {x + \dfrac{b}{{2a}}} \right) = \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \].
Or, taking the constant term into the L.H.S, we get:
\[ \Rightarrow x = - \dfrac{b}{{2a}} \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \].
Now, we can re-write the above equation by following form:
\[ \Rightarrow x = - \dfrac{b}{{\sqrt {4{a^2}} }} \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \]
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{\sqrt {4{a^2}} }}\]
\[ \Rightarrow \] \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]