Worked example
Example
Sketch the graph of \(y = {x^2} + 2x + 3\)
Answer
First we need to complete the square to get the coordinates of the turning point.
\(y = {x^2} + 2x + 3\)
\(y = {(x + 1)^2} - 1 + 3\)
\(y = {(x + 1)^2} + 2\)
Therefore the coordinates of the turning point are (-1, 2).
If we recall the general equation: \(y = a{x^2} + bx + c\) then if:
- a > 0, then the shape of the parabola is like a happy face and the nature shows it is a minimum turning point
- a < 0, then the shape of the parabola is like a sad face and the nature shows it is a maximum turning point
Therefore, as a \(\textgreater 0\) the above equation has a minimum turning point at (-1, 2).
Next, we need to find the roots of the equation. We can use the 'discriminant' to show how many roots there are, if any:
- \({b^2} - 4ac\textgreater0\) means there are two roots
- \({b^2} - 4ac = 0\) means there is one root (because the turning point is on the x-axis)
- \({b^2} - 4ac\textless0\) means there are no roots
\(y = {x^2} + 2x + 3\)
\({b^2} - 4ac\) where \(a = 1,\,b = 2\,and\,c = 3\)
\(= {2^2} - (4 \times 1 \times 3)\)
\(= 4 - 12 = - 8\) which is < 0 therefore there are no roots.
A parabola cuts the y axis, when \(x = 0\):
\(y = {(0)^2} + 2(0) + 3\)
\(y = 3\)
Therefore the y-intercept is (0, 3)
Therefore the graph looks like:
More guides on this topic
- Expansion of brackets
- Factorising an algebraic expression
- Completing the square in a quadratic expression
- Simplifying algebraic fractions
- Applying the four operations to algebraic fractions
- Determining the equation of a straight line
- Working with linear equations and inequations
- Working with simultaneous equations
- Changing the subject of a formula
- Determine the equation of a quadratic function from its graph
- Identifying features of a quadratic function
- Solving a quadratic equation
- Solving a quadratic equation using the quadratic formula
- Using the discriminant to determine the number of roots
- Video playlist
