Find the vertex focus directrix and axis of symmetry of the parabola

[tex]-\left(y+1\right)=\left(x-2\right)^2[/tex][tex]\mathrm{Divide\:both\:sides\:by\:}-1[/tex][tex]\frac{-\left(y+1\right)}{-1}=\frac{\left(x-2\right)^2}{-1}[/tex][tex]y+1=-x^2+4x-4[/tex][tex]\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}[/tex][tex]y+1-1=-x^2+4x-4-1[/tex]

Simplify:

[tex]y=-x^2+4x-5[/tex][tex]\mathrm{The\:parabola\:params\:are:}[/tex][tex]a=-1,\:b=4,\:c=-5[/tex][tex]x_v=-\frac{b}{2a}[/tex][tex]x_v=-\frac{4}{2\left(-1\right)}[/tex]

Remove parentheses:

[tex]=-\frac{4}{-2\cdot \:1}[/tex][tex]\mathrm{Multiply\:the\:numbers:}\:2\cdot \:1=2[/tex][tex]=-\frac{4}{-2}[/tex]

Apply the fraction rule:

[tex]=-\left(-\frac{4}{2}\right)[/tex][tex]\mathrm{Divide\:the\:numbers:}\:\frac{4}{2}=2[/tex][tex]=-\left(-2\right)[/tex][tex]x_v=2[/tex]

Plugging in x_v into the expression:

[tex]y_v=-2^2+4\cdot \:2-5[/tex]

Simplify:

[tex]y_v=-1[/tex][tex]\mathrm{Therefore\:the\:parabola\:vertex\:is}[/tex][tex]\text{ \lbrack VERTEX\rbrack --> }\left(2,\:-1\right)[/tex][tex]\mathrm{If}\:a<0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}[/tex][tex]\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}[/tex][tex]a=-1[/tex][tex]\mathrm{Maximum}\:\left(2,\:-1\right)[/tex]

FOCUS:

[tex]A\:parabola\:is\:the\:locus\:of\:points\:such\:that\:the\:distance\:to\:a\:point\:\left(the\:focus\right)\:equals\:the\:distance\:to\:a\:line\:\left(the\:directrix\right)[/tex][tex]A\:parabola\:is\:the\:locus\:of\:points\:such\:that\:the\:distance\:to\:a\:point\:\left(the\:focus\right)\:equals\:the\:distance\:to\:a\:line\:\left(the\:directrix\right)[/tex][tex]\mathrm{and\:a\:focal\:length\:}\:|p|[/tex][tex]\left(x-2\right)^2=-\left(y+1\right)[/tex]

Switch sides:

[tex]-\left(y+1\right)=\left(x-2\right)^2[/tex][tex]\mathrm{Factor\:}-1[/tex][tex]\left(-1\right)\left(y+\frac{-1}{-1}\right)=\left(x-2\right)^2[/tex]

Simplify:

[tex]\left(-1\right)\left(y+1\right)=\left(x-2\right)^2[/tex][tex]\mathrm{Factor\:}4[/tex][tex]4\cdot \frac{-1}{4}\left(y+1\right)=\left(x-2\right)^2[/tex]

Simplify:

[tex]4\left(-\frac{1}{4}\right)\left(y+1\right)=\left(x-2\right)^2[/tex][tex]\mathrm{Rewrite\:as}[/tex][tex]4\left(-\frac{1}{4}\right)\left(y-\left(-1\right)\right)=\left(x-2\right)^2[/tex][tex]\left(h,\:k\right)=\left(2,\:-1\right),\:p=-\frac{1}{4}[/tex][tex]\mathrm{Parabola\:is\:symmetric\:around\:the\:y-axis\:and\:so\:the\:focus\:lies\:a\:distance\:}p\mathrm{\:from\:the\:center\:}[/tex][tex]\left(2,\:-1\right)\mathrm{\:along\:the\:y-axis}[/tex][tex]=\left(2,\:-1+\left(-\frac{1}{4}\right)\right)[/tex][tex]\mathrm{Refine}[/tex][tex][FOCUS]-->\left(2,\:-\frac{5}{4}\right)[/tex]

Directrix:

[tex]\mathrm{Parabola\:is\:symmetric\:around\:the\:y-axis\:and\:so\:the\:directrix\:is\:a\:line\:parallel\:to\:the\:x-axis,\:a\:distance\:}[/tex][tex]-p\mathrm{\:from\:the\:center\:}\left(2,\:-1\right)\mathrm{\:y-coordinate}[/tex][tex]y=-1-p[/tex][tex]y=-1-\left(-\frac{1}{4}\right)[/tex][tex]\mathrm{Refine}[/tex][tex]y=-\frac{3}{4}[/tex]

Axis of simmetry:

[tex]\mathrm{Parabola\:is\:of\:the\:form\:}4p\left(y-k\right)=\left(x-h\right)^2\mathrm{\:and\:is\:symmetric\:around\:the\:}y\mathrm{-axis}[/tex][tex]\mathrm{Axis\:of\:symmetry\:is\:a\:line\:parallel\:to\:the\:}y\mathrm{-axis\:which\:intersects\:the\:vertex:}[/tex][tex]x=2\text{ \lbrack Axis of simmetry\rbrack}[/tex]