Question 1186906

A parabola has a y-intercept of {{{-4 }}}=>({{{0}}},{{{-4 }}}) =>{{{k=-4}}}

and passes through the points ({{{-2}}}, {{{8}}}) and ({{{1}}}, {{{-1}}}). 

Determine the vertex of the parabola.

use  the vertex form of a quadratic equation: 

{{{y = a ( x- h )^2 + k}}}........use given point ({{{-2}}}, {{{8}}})  and intercept {{{k=-4}}}

{{{8= a ( -2- h )^2 -4}}}......solve for {{{a}}}

{{{(8+4)/( -2- h )^2 = a }}}

{{{12/( -2- h )^2 = a }}}........eq.1


{{{y = a ( x- h )^2 + k}}}........use given point ({{{1}}}, {{{-1}}})  and intercept {{{k=-4}}}

{{{-1 = a (1- h )^2 -4}}}

{{{(-1+4) /(1- h )^2= a  }}}

{{{3 /(1- h )^2= a  }}}...........eq.2

from eq.1 and eq.2 we have

{{{12/( -2- h )^2 = 3 /(1- h )^2}}}..........solve for {{{h}}}

{{{12(1- h )^2 = 3( -2- h )^2}}}.....simplify

{{{4(1- h )^2 = ( -2- h )^2}}}

{{{4(h^2 - 2 h + 1 ) = h^2 + 4 h + 4}}}

{{{4h^2 - 8 h + 4 = h^2 + 4 h + 4}}} 

{{{4h^2 - 8 h + 4 - h^2 - 4 h -4=0}}} 

{{{3h^2 - 12h   =0}}} 

{{{(3h - 12)h   =0}}} 

solutions:

{{{h=0}}}
or
{{{3h - 12   =0}}} =>{{{3h=12}}}=>{{{h=4}}}

the vertex of the parabola is at ({{{0}}},{{{-4}}})  or  ({{{4}}},{{{-4}}}) 


go to {{{12/( -2- h )^2 = a }}}........eq.1, substitute {{{h}}}

if {{{h=0}}}

{{{12/( -2- 0 )^2 = a }}}

{{{12/4 = a }}}

{{{a=3}}}

or

if {{{h=4}}}

{{{12/( -2- 4 )^2 = a }}}

{{{12/36 = a }}}

{{{a=1/3}}}




so, your parabola is:

{{{y = 3 ( x- 0 )^2 -4}}}
{{{y = 3 x^2 -4}}}

or
{{{y = (1/3) ( x- 4 )^2 -4}}}


let's see both solutions on the graph:

1. {{{y = 3 x^2 -4}}}

{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(0,-4,.12), locate(0,-4,V(0,-4)),
circle(1,-1,.12), locate(1,-1,p(1,-1)),
circle(-2,8,.12), locate(-2,8,p(-2,8)),
graph( 600, 600, -10, 10, -10, 10, 3x^2 -4)) }}}


2.

{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(4,-4,.12), locate(4,-4,V(4,-4)),
circle(1,-1,.12), locate(1,-1,p(1,-1)),
circle(-2,8,.12), locate(-2,8,p(-2,8)),
graph( 600, 600, -10, 10, -10, 10, (1/3)( x- 4 )^2 -4)) }}}