\(y = {x^2} - 8x + 13\)
Use the standard form of a quadratic equation with vertex\(\left( {h,k} \right)\),
\(y = a{\left( {x - h} \right)^2} + k\)
Rewrite the given quadratic equation in the standard form by completing the square,
\(\begin{array}{l}y = {x^2} - 8x + 13\\y = {x^2} - 8x + 16 - 16 + 13\\y = {\left( {x - 4} \right)^2} - 3\end{array}\)
Therefore, the standard form of given quadratic equation is,
\({y = {{\left( {x - 4} \right)}^2} - 3}\)
Compare equation (3)\(y = {\left( {x - 4} \right)^2} - 3\) with standard form (2)\(y = a{\left( {x - h} \right)^2} + k\)
\(a = 1,h = 4,k = - 3\)
So, the vertex of the given equation is\(\left( {h,k} \right) = \left( {4, - 3} \right)\) .
Axis of symmetry will be,
\(x = h \Rightarrow x = 4\)
Since, the sign of a is positive,
Hence, the parabola opens “Upwards.”
Tags:
quadratic equation