Sketching a quadratic functionSample questions

\(y = - {x^2} + 2x - 3\)

\(y = - 1[{x^2} - 2x + 3]\)

\(y = - 1[{(x - 1)^2} - 1 + 3]\)

\(y = - 1[{(x - 1)^2} + 2]\)

\(y = - 1{(x - 1)^2} - 2\)

\(a=-1\)

\({b^2} - 4ac\) 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.

Cuts y-axis when \(x = 0\):

\(y = - {(0)^2} + 2(0) - 3\)

\(y = - 3\)

Roots occur when \(y = 0\)

\((x + 1)(x - 3) = 0\)

\((x + 1) = 0\,and\,(x - 3) = 0\)

\(x = - 1\,and\,x = 3\)

If we imagine multiplying out the brackets, you can see we will get a positive \({x^2}\), therefore since a > 0, it's a minimum turning point.

Therefore turning point occurs when \(x = 1\). To find the y coordinate of our turning point, we can substitute this into the original equation.

When \(x = 1\),

\(y = (1 + 1)(1 - 3)\)

\(y = (2)( - 2)\)

\(y = - 4\)

Parabola cuts y axis when \(x = 0\). Therefore, by substituting this in, we get:

\(y = (0 + 1)(0 - 3)\)

\(y = (1)( - 3)\)

\(y = - 3\)