Question

What is the equation of a parabola that goes through `(-4, 1), (2,7)`?

Answer 1

`f(x) = kx^2+(2k+1)x+(5-8k)" "` for any `k != 0`

Explanation:

Let us start with the equation of a line that passes through these two points.

The slope `m` of a line through two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula:

`m = (Delta y)/(Delta x) = (y_2-y_1)/(x_2-x_1) = (7-1)/(2-(-4)) = 6/6 = 1`

Using the point `(2, 7)` we can write the equation of the line in point slope form as:

`y - 7 = m(x - 2) = x - 2`

Add `7` to both ends to get slope intercept form:

`y = x + 5`

To get the equation of a parabola through the same two points, we can add this linear function to a quadratic function that has zeros at `x=-4` and `x=2` to get:

`f(x) = (x+5) + k(x+4)(x-2)`

`color(white)(f(x)) = (x+5) + kx^2+2kx-8k`

`color(white)(f(x)) = kx^2+(2k+1)x+(5-8k)`

Here are a few examples for `k=1`, `k=1/2` and `k=-1/2` plotted along with the line `y = x+5` ...

graph{(y - ((0)x^2+(2(0)+1)x+(5-8(0))))(y - ((1)x^2+(2(1)+1)x+(5-8(1))))(y - ((1/2)x^2+(2(1/2)+1)x+(5-8(1/2))))(y - ((-1/2)x^2+(2(-1/2)+1)x+(5-8(-1/2)))) = 0 [-11.24, 8.76, -6, 9.5]}