# 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 ?

Aug 13, 2016

$m = 1$

#### Explanation:

$f \left(x\right) = {x}^{2} + 2 \left(m + 1\right) x + m + 3$

$= {\left(x + \left(m + 1\right)\right)}^{2} - {\left(m + 1\right)}^{2} + m + 3$

$= {\left(x + \left(m + 1\right)\right)}^{2} - {m}^{2} - m + 2$

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:

$0 = - {m}^{2} - m + 2 = \left(2 + m\right) \left(1 - m\right)$

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

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

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

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Alternative method

$f \left(x\right) = {x}^{2} + 2 \left(m + 1\right) x + m + 3$

This is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 2 \left(m + 1\right)$, $c = \left(m + 3\right)$

It has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(2 \left(m + 1\right)\right)}^{2} - 4 \left(1\right) \left(m + 3\right)$

$= 4 \left({m}^{2} + 2 m + 1 - m - 3\right)$

$= 4 \left({m}^{2} + m - 2\right) = 4 \left(m + 2\right) \left(m - 1\right)$

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 $f \left(x\right) = {x}^{2} - 2 x + 1 = {\left(x - 1\right)}^{2}$, which is zero when $x = 1$. This is on the positive part of the $x$ axis.

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