How to solve Use completing the square to find the vertex of the following parabolas
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To use completing the square to find the vertex of the given parabola, we proceed as follows:
[tex]g(x)=x^2-5x+14[/tex]- we divide the coefficient of x by 2 and add and subtract the square of the result, as follows:
[tex]g(x)=x^2-5x+(\frac{5}{2})^2-(\frac{5}{2})^2+14[/tex]- simplify the expression as follows:
[tex]\begin{gathered} g(x)=(x^2-5x+(\frac{5}{2})^2)-(\frac{5}{2})^2+14 \\ \end{gathered}[/tex][tex]g(x)=(x^{}-\frac{5}{2})^2-(\frac{5}{2})^2+14[/tex][tex]g(x)=(x^{}-\frac{5}{2})^2-\frac{25}{4}^{}+14[/tex][tex]g(x)=(x^{}-\frac{5}{2})^2-\frac{25}{4}^{}+\frac{56}{4}[/tex][tex]g(x)=(x^{}-\frac{5}{2})^2+\frac{-25+56}{4}^{}[/tex][tex]g(x)=(x^{}-\frac{5}{2})^2+\frac{31}{4}^{}[/tex]From the general vertex equation, given as:
[tex]g(x)=a(x-h)^2+k[/tex]The coordinate of the vertex is taken as: (h, k)
Therefore, given:
[tex]g(x)=(x^{}-\frac{5}{2})^2+\frac{31}{4}^{}[/tex]We have the vertex to be:
[tex](\frac{5}{2},\frac{31}{4})\text{ or (2.5, 7.75)}[/tex]