Answer
a. $y=-4x^2+50x$.
b. $y=-4(6)^2+50(6)$,
$y=156$
The car travels $156$ feet $6$ seconds after the brakes are applied
Work Step by Step
a.To write the quadratic equation $y=ax^2+bx+c$, first we need to get $a,b$,and $c$ by solving the mathematical model that describes the relationship between the number of feets a car travels once the brakes are applied.
The mathematical model is,
$x=1$, $y=46$ { $46=a+b+c$,
$x=2$, $y=84$ { $84=4a+2b+c$,
$x=3$, $y=114$ { $114=9a+3b+c$,
let's take a pair of two equations and eliminate $c$,
{$46=a+b+c$ and {$84=4a+2b+c$
{$84=4a+2b+c$ , { $114=9a+3b+c$ ,
multiply one of pair's of equation by $-1$, to eliminate $c$
{$46=a+b+c$
$-1${{$84=4a+2b+c$, $->$ $-84=-4a-2b-c$
Adding the equation will result in the first pair of the equation to be, $-38=-3a-b$.
doing the same to the second pair of the equation will result,
$30=5a+b$
Arrange the result to eliminate the one variable and solve for one,
{$-38=-3a-b$
{$30=5a+b$,
adding the equation,
$(-38+30)=(-3+5)a+(-1+1)b$,
$-8=2a$,
$a=-4$,
back-substituting and solving for $b$,
$30=5(-4)+b$,
$30+20=b$,
$50=b$,
$50=b$.
back-substituting and solving for $c$,
$46=-4+50+c$,
$46-46=c$,
$0=c$.
Therefore the equation will be,
$y=-4x^2+50x$.
b.
$y=-4(6)^2+50(6)$,
$y=-144+300$,
$y=156$
The car travels 156 feet 6 seconds after the brakes are applied
