Question

Question #a26d7

Answer 1

`y=+1.15x^2+3.85x-2`

Explanation:

The quadratic is a parabolic equation. One of its standardised forms is:

`y=ax^2+bx+c`

We are given 3 points for `(x,y)`
There are 3 unknowns `{a,b,c}`

So we have 3 unknowns and 3 equations thus solvable.

Point A
`P_A-> (1,3) ->3=a(1)^2+b(1)+c" "..............Equation(1)`

Point B
`P_B->(-4,1)->1=a(-4)^2+b(-4)+c" "..Equation(2)`

Point C
`P_C->(0,-2)->-2=a(0)^2+b(0)+c" ".......Equation(3)`
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`color(blue)("Determining the value of "c)`

Consider equation(1) : this make life a bit easier as we can directly read off the value of `c`

`color(green)(-2=0+0+c => c=-2)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determining the value of "b)`

Substituting for c we have:

`3=a(1)^2+b(1)-2" "..............Equation(1)`
`1=a(-4)^2+b(-4)-2" "......Equation(2)`

`3=a+b-2" "..............Equation(1)`
`1=16a-4b-2" ".........Equation(2)`

To find b we need to get rid of a so multiply equation (1) by 16 and then subtract

`48=16a+16b-32" ".....................Equation(1_a)`
`ul(color(white)(4)1=16a-color(white)(1)4b-2)" "larr" subtract " Equation(2)`
`47=color(white)(.)0color(white)(.)+20b-30`

`color(green)(b=(47+30)/20=3.85)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determining the value of "a)`
Substitute for b and c in equation (say ) 1: looks easiest

`3=a+b+c" "->3=a+3.85-2`

`color(green)(a=+1.15)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`y=+1.15x^2+3.85x-2`