Worked example

Sketch the graph of \(y = {x^2} + 2x + 3\)

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:

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