Question

How do you write an equation of the quadratic function whose graph contains the points (-4,-57), (2, 39), (5, 168)?

Answer 1

`3x^2+6x+3`

Explanation:

General form of a quadratic equation is
`color(white)("XX")y=ax^2+bx+c`

So for the given points, we have
[1]`color(white)("XX")-57=a(-4)^2+b(-4)+c`
[2]`color(white)("XX")39=a(2)^2+b(2)+c`
[3]`color(white)("XX")168=a(5)^2+b(5)+c`

Simplifying these:
[3]`color(white)("XX")16a-4b+c=-57`
[4]`color(white)("XX")4a+2b+c=39`
[5]`color(white)("XX")25a+5b+c=168`

Subtracting [4] from [3]
[6]`color(white)("XX")12a-6b= -96`
Subtracting [4] from [5]
[7]`color(white)("XX")21a+3b=129`

Multiplying [7] by 2
[8]`color(white)("XX")42a+6b=258`

Adding [6] and [8]
[9]`color(white)("XX")54a = 162`

Dividing [9] by 3
[10]`color(white)("XX")a=3`

Substituting `3` for `a` in [6]
[11]`color(white)("XX")12(3)-6b=-96`

Simplifying
[12]`color(white)("XX")-6b =-132`

Dividing [12] by `-6`
[13]`color(white)("XX")b=12`

Substituting `12` for `b` (from [13]) and `3` for `a` (from [10]) in [4]
[14]`color(white)("XX")4(3)+2(12)+c = 39`

Simplifying
[15]`color(white)("XX")12+24+c=39`

[16]`color(white)("XX")c=3`

Answer 2

`y=3x^2+22x-17`

Explanation:

First, we will create three equations, one for each point. It will be in the general form `y=ax^2+bx+c`.

First Equation from P(-4,-57)
`y=ax^2+bx+c`
`(-57)=a(-4)^2+b(-4)+c`
`-57=16a-4b+c`

Second Equation from P(2,39)
`y=ax^2+bx+c`
`(39)=a(2)^2+b(2)+c`
`39=4a+2b+c`

Third Equation from P(5,168)
`y=ax^2+bx+c`
`(168)=a(5)^2+b(5)+c`
`168=25a+5b+c`

System of Equations:
`-57=16a-4b+c`
`39=4a+2b+c`
`168=25a+5b+c`

Now, we can solve for the values of `a`, `b`, and `c`. Let's try getting two more equations to help us solve this system.

Fourth Equation taken from the first and second equations:
We will first get the value of `c` in the first equation then we will substitute it into `c` in the second equation.
`-57=16a-4b+c`
`-57-16a+4b=16a-4b+c-16a+4b`
`-57-16a+4b=c`

`39=4a+2b+c`
`39=4a+2b+(-57-16a+4b)`
`39=4a+2b-57-16a+4b`
`39+57=-12a+6b`
`96=-12a+6b`

Let's simplify this to:
`16=-2a+b`

Fifth Equation taken from the first and third equations:
We will substitute the value of `c` in the first equation into `c` in the third equation.
`168=25a+5b+c`
`168=25a+5b+(-57-16a+4b)`
`168=25a+5b-57-16a+4b`
`168+57=9a+9b`
`225=9a+9b`

Let's simplify this to:
`25=a+b`

Solution
Now, we will use the 4th and 5th equation to solve for `a` and `b`. I'll use the elimination method since it is easier:

`-2a+b=16`
`ul(-a-b=-25)`
`-3a=-9`
`a=3`

`25=a+b`
`25=3+b`
`b=22`

To solve for `c` just substitute the values of `a` and `b` into any of the first three equations.
`39=4a+2b+c`
`39=4(3)+2(22)+c`
`39=12+44+c`
`c=-17`

So the values of `a`, `b`, and `c` are:
`a=3`
`b=22`
`c=-17`

Finally, to get the quadratic equation, just substitute the values of `a`, `b`, and `c` into the general form `y=ax^2+bx+c`.
`y=ax^2+bx+c`
`color(red)(y=3x^2+22x-17)`