Given that $f(x) - g(x) = 3ax^2 + 2(a+1)x + a+1$, we need to find the range of values for $a$ such that $f(x) > g(x)$ for all real numbers $x$. Then, for a specific value of $a$, we need to find the coordinates of the intersection point of the two parabolas $y = f(x)$ and $y = g(x)$.

1. Problem Description

Given that $f(x) - g(x) = 3ax^2 + 2(a+1)x + a+1$, we need to find the range of values for $a$ such that $f(x) > g(x)$ for all real numbers $x$.

Then, for a specific value of $a$, we need to find the coordinates of the intersection point of the two parabolas $y = f(x)$ and $y = g(x)$.

2. Solution Steps

Since $f(x) > g(x)$ for all $x$, we have $f(x) - g(x) > 0$ for all $x$.

This means that $3ax^2 + 2(a+1)x + a+1 > 0$ for all $x$.

For a quadratic $Ax^2 + Bx + C > 0$ for all $x$, we need $A > 0$ and the discriminant $B^2 - 4AC < 0$.

In this case, $A = 3a$, $B = 2(a+1)$, and $C = a+1$.

So, we require $3a > 0$ and $(2(a+1))^2 - 4(3a)(a+1) < 0$.

From $3a > 0$, we get $a > 0$.

The discriminant condition is:

$4(a+1)^2 - 12a(a+1) < 0$

$4(a^2 + 2a + 1) - 12a^2 - 12a < 0$

$4a^2 + 8a + 4 - 12a^2 - 12a < 0$

$-8a^2 - 4a + 4 < 0$

$8a^2 + 4a - 4 > 0$

$2a^2 + a - 1 > 0$

$(2a-1)(a+1) > 0$

This inequality holds when $a > \frac{1}{2}$ or $a < -1$.

Since we also have $a > 0$, the solution is $a > \frac{1}{2}$.

Therefore, the inequality $f(x) > g(x)$ holds for all real numbers $x$ when $a > \frac{1}{2}$.

Now, we consider the case when $a = \frac{1}{2}$.

Then, $f(x) - g(x) = 3(\frac{1}{2})x^2 + 2(\frac{1}{2}+1)x + \frac{1}{2}+1 = \frac{3}{2}x^2 + 3x + \frac{3}{2} = \frac{3}{2}(x^2 + 2x + 1) = \frac{3}{2}(x+1)^2$.

If $a = \frac{1}{2}$, then $f(x) - g(x) = 0$ when $x = -1$.

In this case, $f(x) = g(x)$ when $x=-1$.

Thus, $x = -1$ is the x-coordinate of the intersection point.

Also when $a = 1/2$, $f(x) - g(x) = 3a x^2 + 2(a+1)x + a+1 = 0$. So, $f(x) = g(x)$.

If $x = -1$, $f(x) = g(x)$. Since $f(x) - g(x) = 3ax^2 + 2(a+1)x + a+1$, $f(x) = g(x)$. Let $a = 1/2$. Then

$f(x) - g(x) = 3/2x^2 + 3x + 3/2 = 3/2 (x^2 + 2x + 1) = 3/2 (x+1)^2 = 0$ when $x = -1$.

So $f(x) = g(x)$ when $x = -1$. Let $f(x) = ax^2 + bx + c$ and $g(x) = dx^2 + ex + f$.

$f(-1) - g(-1) = 0$.

$f(-1) = g(-1)$.

$3/2x^2 + 3x + 3/2 = 0$. $3/2 (-1)^2 + 3 (-1) + 3/2 = 3/2 -3 + 3/2 = 0$.

So, when $a = 1/2$, $f(x) = g(x)$ at $x = -1$.

$f(x)-g(x) = 3/2 (x+1)^2$.

Let us assume that the vertex of $g(x)$ is $(0, 0)$. $g(x) = x^2$ and $f(x) = 5/2 x^2 + 3x + 3/2$.

When $a = 1/2$, the graph $f(x) - g(x) = 3/2 (x+1)^2$ has a root $-1$.

Since we are given that $a = \frac{1}{2}$, the $x$ value is $x = -1$.

From $f(x) - g(x) = \frac{3}{2}(x+1)^2$, we know that $f(x)$ and $g(x)$ are tangent to each other when $x=-1$.

Assume $f(x) = ax^2 + bx + c$ and $g(x) = dx^2 + ex + f$.

Given $f(x) - g(x) = \frac{3}{2}(x+1)^2 = \frac{3}{2}(x^2+2x+1)$.

Let $f(x) - g(x) = \frac{3}{2}x^2 + 3x + \frac{3}{2}$.

If $g(x) = 0$, then $f(x) = \frac{3}{2}x^2 + 3x + \frac{3}{2}$.

Then, $f(-1) = \frac{3}{2}(-1)^2 + 3(-1) + \frac{3}{2} = \frac{3}{2} - 3 + \frac{3}{2} = 0$.

The intersection point is $(-1, 0)$.

3. Final Answer

$a > \frac{1}{2}$. The symbol is $>$, which is option ②. Thus O=

2. $P=1$, $Q=2$. Thus $a > \frac{1}{2}$.

When $a = \frac{1}{2}$, the intersection point is $(-1, 0)$.

So $R = -1$, $S=1$, $T=0$, $U=0$, $V=1$. The coordinate of the point is $(-1, 0)$.

Final Answer: $a>1/2$, $O=2, P/Q=1/2$, $(-1, 0)$

O=2

P=1

Q=2

R=-1

S=1

T=0

U=0

V=1

Final Answer: a ② 1/2 である。また、a = 1/2 のとき、2つの放物線y = f(x), y = g(x) の共有点の座標は ( -1/1 , 0/1 ) である。