Question

I have no idea how to do this stuff.

Answer 1

Answer:
`y=2x^(2)-5x+7`

Explanation:

The quadratic equation in its standard form is:

`y=ax^(2)+bx+c`

Now, we are given 5 points of the parabola (if you graph a quadratic equation you will have a parabola), however we only need to choose three points to find the coeficients
`a`,
`b` and
`c` in the quadratic equation.

So, let's choose the first three points:

(-1,14):

`14=a(-1)^(2)+b(-1)+c`

`14=a-b+c` (1)

(0,7):

`7=a(0)^(2)+b(0)+c`

`7=c` (2)

(1,4):

`y=2x^(2)-5x+7`

`4=a(1)^(2)+b(1)+c`

`4=a+b+c` (3)

Substituting (2) in (1) and (3):

`14=a-b+7` (4)

`4=a+b+7` (5)

At this point we have a system with two equations.

Adding (4) to (5):

`18=2a+14` (6)

Isolating
`a`:

`a=2` (7)

Substituitng (7) in (3):

`4=2+b+7` (8)

Isolating
`b`:

`b=-5` (9)

Now we have the three coeficients and we can write the quadratic equation:

`y=2x^(2)-5x+7`