This calculator finds x and y intercepts, vertex, focus and plots the quadratic function.
The calculator will generate a detailed explanation for each of these computations.

Quadratic function has the form $ f(x) = ax^2 + bx + c $ where a, b and c are numbers
You can sketch quadratic function in 4 steps. I will explain these steps in following examples.
Example 1:
Sketch the graph of the quadratic function
$$ {\color{blue}{ f(x) = x^2+2x3 }} $$Solution:
In this case we have $ a=1, b=2 $ and $c=3$
STEP 1: Find the vertex.
To find x  coordinate of the vertex we use formula:
$$ x=\frac{b}{2a} $$So, we substitute $1$ in for $a$ and $2$ in for $b$ to get
$$ x=\frac{b}{2a} = \frac{2}{2\cdot1} = 1 $$To find y  coordinate plug in $x=1$ into the original equation:
$$ y = f(1) = (1)^2 + 2\cdot(1)  3 = 1  2  3 = 4 $$So, the vertex of the parabola is $ {\color{red}{ (1,4) }} $
STEP 2: Find the yintercept.
To find y  intercept plug in $x=0$ into the original equation:
So, the yintercept of the parabola is $ {\color{blue}{ y = 3 }} $
STEP 3: Find the xintercept.
To find x  intercept solve quadratic equation $f(x)=0$ in our case we have:
$$ x^2+2x3 = 0 $$Solutions for this equation are:
$$ {\color{blue}{ x_1 = 3 }} ~~~\text{and}~~~ {\color{blue}{ x_2 = 1 }} $$( to learn how to solve quadratic equation use quadratic equation solver )
STEP 4: plot the parabola.
Example 2:
Sketch the graph of the quadratic function
$$ {\color{blue}{ f(x) = x^2+2x2 }} $$Solution:
Here we have $ a=1, b=2 $ and $c=2$
The xcoordinate of the vertex is:
$$ {\color{blue}{ x = \frac{b}{2a} }} = \frac{2}{2\cdot(1)}= 1 $$The ycoordinate of the vertex is:
$$ y = f(1) = 1^2+2\cdot12 = 1 + 2  2 = 1 $$The yintercept is:
$$ y = f(0) = 0^2+2\cdot02 = 0 + 0  2 = 2 $$In this case xintercept doesn't exist since equation $x^2+2x2=0$ does not has the solutions (use quadratic equation solver to check ). So, in this case we will plot the graph using only two points
Please tell me how can I make this better.
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