Question

How many quadratic, cubic and quartic functions are there with unique `x` intercept `(-3, 0)` and passing through `(6, 23)`?

Answer 1

See explanation...

Explanation:

How many quadratic, cubic and quartic functions are there with unique `x` intercept `(-3, 0)` and passing through `(6, 23)`?

1. Quadratic

If a quadratic function only has one `x`-intercept, then that must be where its vertex is.

So a suitable quadratic function with `x`-intercept at `(-3, 0)` can be written in the form:

`f(x) = a(x+3)^2`

for some constant `a` to be determined.

Since we want this to pass through `(6, 23)`, those `x` and `y` values must satisfy the equation:

`color(blue)(23) = a(color(blue)(6)+3)^2 = 81a`

So `a=23/81` and we can write our function as:

`f(x) = 23/81(x+3)^2`

So there is precisely one quadratic function satisfying the conditions.

2. Cubic

Unlike quadratics, cubic functions can easily have just one `x`-intercept. In our example, the function must have a linear factor `(x+3)` in order that it has `(-3, 0)` as an `x`-intercept. The remaining quadratic must have only non-real zeros, which we can arrange by writing:

`f(x) = k(x+3)((x+a)^2+b)`

where `k != 0` is a real constant, `a` is a real constant and `b > 0` is a positive real constant.

To pass through `(6, 23)` those `x` and `y` coordinates must satisfy this equation. So:

`color(blue)(23) = k(color(blue)(6)+3)((color(blue)(6)+a)^2+b)`

`color(blue)(23) = 9k((a+6)^2+b)`

So for any `a in RR` and `b > 0` we can set:

`k = 23/(9((a+6)^2+b))`

Then a cubic with the desired properties is:

`f(x) = 23/(9((a+6)^2+b))(x+3)((x+a)^2+b)`

There are uncountably infinitely many such cubics.

3. Quartic

In order to have just one `x`-intercept, we require our function to have a double zero by having a factor `(x+3)^2`

Similar to the cubic case, we can then write:

`f(x) = k(x+3)^2((x+a)^2+b)`

where `a in RR` and `b > 0`

and choose:

`k = 23/(81((a+6)^2+b))`

So there are uncountably infinity many suitable quartics too.