Question

The parabola `f(x) = x^2 + 2(m+1)x + m + 3` is tangent to `x` axis in the negative side. What is the value of `m` ?

Answer 1

`m=1`

Explanation:

`f(x) = x^2+2(m+1)x+m+3`

`=(x+(m+1))^2-(m+1)^2+m+3`

`=(x+(m+1))^2-m^2-m+2`

This is in vertex form with intercept `-m^2-m+2`

In order for it to be tangential to the `x` axis the intercept must be `0`. That is:

`0 = -m^2-m+2 = (2+m)(1-m)`

So `m = -2` or `m = 1`.

If `m = -2` then the vertex is at `(1, 0)` on the positive part of the `x` axis.

If `m = 1` then the vertex is at `(-2, 0)` on the negative part of the `x` axis, as required.

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Alternative method

`f(x) = x^2+2(m+1)x+m+3`

This is of the form `ax^2+bx+c` with `a=1`, `b=2(m+1)`, `c=(m+3)`

It has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (2(m+1))^2-4(1)(m+3)`

`=4(m^2+2m+1-m-3)`

`=4(m^2+m-2) = 4(m+2)(m-1)`

If the parabola is tangential to the `x` axis then it has a double root and we require `Delta = 0`.

Hence `m=-2` or `m=1`

If `m=-2` then `f(x)=x^2-2x+1 = (x-1)^2`, which is zero when `x=1`. This is on the positive part of the `x` axis.

If `m=1` then `f(x)=x^2+4x+4=(x+2)^2`, which is zero when `x=-2`. This is on the negative part of the `x` axis as required.