Question

Given (2,1),(0,-2),(-2,-3), what is the coordinate of the y intercept, the slope and equation of the line?

Answer 1

Assuming the line is a standard (non-rotated) parabola
it has the form:
`y=ax^2+bx+c`

Using the given points `(2,1), (0,-2), (-2,-3)`
we can write three equations in three unknowns (`a,b,c`) and solve for the parabola's coefficients.

1. `1=4a +2b +c`
2. `-2 =0a +0b +c`
3. `-3 = 4a-2b+c`

From equation 2 `c=-2`
and by inspection or standard operations it follows that
`b=1` and `a=1/4`

The equation of the parabola is therefore
`y = 1/4x^2+x-2`

We were told the y-intercept (it's the point where `x=0`)
The y-intercept is `-2`

We are not given a point at which to evaluate the slope (I'm assuming it is the slope of the tangent that is being asked for)
so the best we can do is give the general formula for the slope at a point `x`
namely the derivative of `y` with respect to `x`
`1/2x+1`