This calculator graphs the quadratic function of the form f(x)=ax2+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.
Graph $ \color{blue}{ f(x) = x^2} $
solution
Step 1:
X - intercept is:
$$ \color{blue}{ x_1 = 0 } $$
To find the x-intercepts, we need to solve equation $ x^2 = 0 $. (use the quadratic equation solver to view a detailed explanation of how to solve the equation)
Step 2:
Y - intercept is point: $ y-inter=\left(0,~0\right) $
To find y - coordinate of y - intercept, we need to compute $ f(0) $. In this example we have:
$$ f(\color{blue}{0}) = 1 \cdot \color{blue}{0}^2 + 0 \cdot \color{blue}{0} + 0 = 0$$
Step 3:
Vertex is point: $V=\left(0,~0\right) $
To find the x - coordinate of the vertex we use formula:
$$ x = -\frac{b}{2a} $$
In this example: $ a = 1, b = 0, c = 0 $. So, the x-coordinate of the vertex is:
$$ x = -\frac{b}{2a} = -\frac{ 0 }{ 2 \cdot 1 } = 0 $$$$ y = f \left( 0 \right) = 1 \left( 0 \right)^2 + 0 \cdot 0 ~ + ~ 0 = 0 $$
Step 4:
Focus is point: $ F=\left(0,~\dfrac{ 1 }{ 4 }\right)$
The x - coordinate of the focus is $ x = -\dfrac{b}{2a} $
The y - coordinate of the focus is $ y = \dfrac{1-b^2}{4a} + c $
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Examples
ex 1:
Graph quadratic function: y=x2/2-3x-4.
ex 2:
Graph the function f(x)=x2+0.5x-2/3.
ex 3:
Graph the function f(x)=- x2+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:
$$ f(0) = (0)^2 + 2\cdot(0) - 3 = 0 - 0 - 3 = -3 $$
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
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