Question

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

Answer 1

Step-by-step explanation

`\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-(b)/(2a)`

Isolating y from the expression:

Switching sides:

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

Simplify:

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

Remove parentheses:

`=-(4)/(-2\cdot \:1)`
`\mathrm{Multiply\:the\:numbers:}\:2\cdot \:1=2`
`=-(4)/(-2)`

Apply the fraction rule:

`=-\left(-(4)/(2)\right)`
`\mathrm{Divide\:the\:numbers:}\:(4)/(2)=2`
`=-\left(-2\right)`
`x_v=2`

Plugging in x_v into the expression:

`y_v=-2^2+4\cdot \:2-5`

Simplify:

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

FOCUS:

`\mathrm{and\:a\:focal\:length\:}\:|p|`
`\left(x-2\right)^2=-\left(y+1\right)`

Switch sides:

`-\left(y+1\right)=\left(x-2\right)^2`
`\mathrm{Factor\:}-1`
`\left(-1\right)\left(y+(-1)/(-1)\right)=\left(x-2\right)^2`

Simplify:

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

Simplify:

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

Directrix:

`-p\mathrm{\:from\:the\:center\:}\left(2,\:-1\right)\mathrm{\:y-coordinate}`
`y=-1-p`
`y=-1-\left(-(1)/(4)\right)`
`\mathrm{Refine}`
`y=-(3)/(4)`

Axis of simmetry:

`\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}`
`\mathrm{Axis\:of\:symmetry\:is\:a\:line\:parallel\:to\:the\:}y\mathrm{-axis\:which\:intersects\:the\:vertex:}`
`x=2\text{ \lbrack Axis of simmetry\rbrack}`