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QuestionWrite the equation of the
parabola $y=-x^{2}-10x-5$ in the
standard form. Then determine the
vertex,
focus,
directrix and the
axis of symmetry of the
parabola. Also, state which way the parabola opens.
Answer$y=-x^{2}-10x-5$
$y=-1\left(x^{2}+10x\right)-5$
$y=-1\left(x^{2}+10x+25\right)+25-5$
$y=-1\left(x+5\right)^2+20$
$y-20= -1\left(x+5\right)^2$
The
parabola opens down words.
The
vertex is given by $V\left(-5,20\right)$
The
focus is given by $F\left(-5,\frac{79}{4}\right)$
The
directrix is given by $y= \frac{81}{4}$
The
axis of symmetry is given by $x=-5$