Sanai Potter

2023-03-13

How to write $f\left(x\right)=-{x}^{2}+6x-13$ into vertex form?

lilcliffordt42

Beginner2023-03-14Added 1 answers

$\text{the equation of a parabola in}\phantom{\rule{1ex}{0ex}}{\text{vertex form}}$ is

$\overline{\underline{\left|{\frac{2}{2}}{y=a{(x-h)}^{2}+k}{\frac{2}{2}}\right|}}$

where an is a constant and the vertex's coordinates are (h, k).

$\text{using the method of}\phantom{\rule{1ex}{0ex}}{\text{completing the square}}$

$f\left(x\right)=-({x}^{2}-6x+13)$

${f\left(x\right)}=-({x}^{2}-6x+9-9+13)$

${f\left(x\right)}=-({(x-3)}^{2}+4)$

$f\left(x\right)}=-{(x-3)}^{2}-4\leftarrow {\phantom{\rule{1ex}{0ex}}\text{in vertex form}$

$\overline{\underline{\left|{\frac{2}{2}}{y=a{(x-h)}^{2}+k}{\frac{2}{2}}\right|}}$

where an is a constant and the vertex's coordinates are (h, k).

$\text{using the method of}\phantom{\rule{1ex}{0ex}}{\text{completing the square}}$

$f\left(x\right)=-({x}^{2}-6x+13)$

${f\left(x\right)}=-({x}^{2}-6x+9-9+13)$

${f\left(x\right)}=-({(x-3)}^{2}+4)$

$f\left(x\right)}=-{(x-3)}^{2}-4\leftarrow {\phantom{\rule{1ex}{0ex}}\text{in vertex form}$