Math Calculators, Lessons and Formulas

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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.

Enter quadratic function of the form (y = ax2 + bx + c )
You can enter either integers (10), decimal numbers(10.12) or FRACTIONS (10/3). Important: The form will NOT let you enter wrong characters (like *, (, ), x, p,...) How to input??

0 1 2 3 4 5 6 7 8 9 - / . del
Show me an explanation.

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:

$$f(0) = (0)^2 + 2\cdot(0) - 3 = 0 - 0 - 3 = -3$$

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