Graphical Representation of Quadratic Equation - Practice Questions & MCQ

We have y = f(x) = ax² + bx + c where a, b, c ∈ R and a ≠ 0

Expression $\mathrm{y}=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=\mathrm{f}(\mathrm{x})$ can be represented as

$
y=a\left(x^2+\frac{b}{a} x+\frac{c}{a}\right)
$

which on further simplification is converted into the form of

$
\begin{aligned}
& \Rightarrow y=a\left[\left(x+\frac{b}{2 a}\right)^2-\frac{D}{4 a^2}\right] \\
& \text { or }\left(y+\frac{D}{4 a}\right)=a\left(x+\frac{b}{2 a}\right)^2
\end{aligned}
$
Now, let $y+\frac{D}{4 a}=Y$ and $x+\frac{b}{2 a}=X$

$
\therefore \mathrm{Y}=\mathrm{aX}^2
$

The shape of the y = f(x) will be parabolic 

Vertex of the parabola will be $\left(\frac{-\mathrm{b}}{2 \mathrm{a}}, \frac{-\mathrm{D}}{4 \mathrm{a}}\right)$ $\left[y+\frac{D}{4 a}=0 \Rightarrow y=-\frac{D}{4 a}\right]$

If the parabola opens upward (when a > 0) then the y value of the vertex represents the least value of the equation, and if opens downward (when a < 0) then the y value of the vertex represents the greatest value of the parabola. 

Both least and greatest values are attained at the x value of the vertex of the parabola 

Hence the graph of any general quadratic equation will look like the below graph (given a>0)

In the general quadratic equation if  $y=a x^2+b x+c=f(x)$ and if a > 0 

Then the parabola opens upward. As given below,

if a < 0 it opens downward. As given below,

$x$-coordinate of the point of intersection of a graph $y=f(x)$ and the $x$-axis gives the root of equation $f(x)=0$. If a graph intersects the $x$-axis at $(p, 0)$, then for $x=p, y=0$, so $x=p$ is a root of the equation $f(x)=0$ (or we can also say $\mathrm{y}=0$ ).

Now, if a, b, c are real numbers

Case 1: If D > 0

We have learnt that if D > 0 and a,b,c are real numbers, then quadratic equations have two distinct real roots. Hence the equation has two different real values of x satisfying the equation, so the graph of the equation must cut the x-axis at two different points, irrespective of the sign of a ( a > 0 or a < 0 ).

So, If $y=\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0=\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{x}=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}}$

Case 2: If  D = 0

We have equal and real roots. Since both roots will coincide, hence the graph will touch the x-axis at one point only, and it will open upward or downward depending upon the value of ‘a’. The graph is shown below. 

Case 3: If D < 0

We know that quadratic equations will have non-real roots which will exist in pairs. So The graph will not cut the x-axis as any real x will not satisfy the equation. So the graph of such an equation will be as given below.

Intersection with y-axis:

$
y=f(x)=a x^2+b x+c
$
For intersection with the $y$-axis, substitute $x=0$ in $y=f(x)=a x^2+b x+c$ we get $\mathrm{y}=\mathrm{c}$.

For example,

Curve $\mathrm{y}=\mathrm{x}^2+\mathrm{x}+1$, cuts the y -axis at $\mathrm{y}=1$ (by putting $\mathrm{x}=0$ in the given equation). So the graph will pass through $(0,1)$. This si the point of intersection of graph with the axis