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- Quadratic Plotter

This calculator graphs the quadratic function of the form f(x)=ax^{2}+bx+x.
The solver also finds the x and y intercepts,
vertex and focus of a quadratic function.
Calculator shows all the work and provides detailed explanation for each step.

working...

Examples

ex 1:

Graph quadratic function: y=x^{2}/2-3x-4.

ex 2:

Graph the function f(x)=x^{2}+0.5x-2/3.

ex 3:

Graph the function f(x)=- x^{2}+5/4.

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Find more worked-out examples in our database of solved problems..

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

$$ {\color{blue}{ f(x) = x^2+2x-3 }} $$**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 y-intercept. **

To find y - intercept plug in $x=0$ into the original equation:

So, the y-intercept of the parabola is $ {\color{blue}{ y = -3 }} $

**STEP 3: Find the x-intercept. **

To find x - intercept solve quadratic equation $f(x)=0$ in our case we have:

$$ x^2+2x-3 = 0 $$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

$$ {\color{blue}{ f(x) = -x^2+2x-2 }} $$**Solution: **

Here we have $ a=-1, b=2 $ and $c=-2$

The x-coordinate of the vertex is:

$$ {\color{blue}{ x = -\frac{b}{2a} }} = -\frac{2}{2\cdot(-1)}= 1 $$The y-coordinate of the vertex is:

$$ y = f(1) = -1^2+2\cdot1-2 = -1 + 2 - 2 = -1 $$The y-intercept is:

$$ y = f(0) = -0^2+2\cdot0-2 = -0 + 0 - 2 = -2 $$In this case **x-intercept doesn't exist** since equation $-x^2+2x-2=0$ does not
has the solutions (use
quadratic equation solver
to check ). So, in this case we will plot the graph using only two points

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