Find the coordinates of the points on $y=-(x+1)^2+4$ that have a distance of $\sqrt {14}$ to $(-1,2)$ Create a function that gives the distance between the point $(-1,2)$  the graph of $$y=-(x+1)^2+4.$$ Find the coordinates of the points on the curve that have a distance of $\sqrt {14}$ units from the point $(-1,2)$.  
I know that the $x$-intercepts are $x=1$ and $x=-3$, and that the vertex is $(-1,4)$. I'm trying to use the distance formula by equating $\sqrt{14}$ to the function, but that is not getting me anywhere. 
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$
 A: The locus of points whose distance from $(-1,2)$ is $\sqrt{14}$ is the circle
$(x+1)^2+(y-2)^2=14$
that is 
$\color{red}{x^2+2 x}+y^2-4 y-9=0$
The parabola equation can be written as $\color{red}{x^2+2 x}=\color{blue}{3-y}\quad(*)$
Therefore the intersection points, the points of the parabola which are $\sqrt{14}$ from the given point, can be found solving
$\color{blue}{3-y} +y^2-4 y-9=0$
which gives $y_1=-1;\;y_2=6$ 
plugging  back $y_1$ in $(*)$
$x^2+2x=4$ gives $x=-1\pm\sqrt 5$
while the other gives no real solutions
Hope this helps

A: You need to solve $$\sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}=\sqrt{14}.$$ Squaring both parts and substituting $u=(x+1)^2$ we get a quadratic equation $u^2-3u-10=0$ which has two roots, $u=5$ and $u=-2$. Only $(x+1)^2=5$ is possible for real numbers so we get $x=\sqrt5-1$ and $x=-\sqrt5-1$
A: this function is given by $$d=\sqrt{(x+1)^2+((x+1)^2-2)^2}$$
simplifying the radicand we have $$d=\sqrt{3\,{x}^{2}-2\,x+2+{x}^{4}+4\,{x}^{3}}$$
defining $$d^2=3\,{x}^{2}-2\,x+2+{x}^{4}+4\,{x}^{3}$$ so we find the first derivative as $$6\,x-2+4\,{x}^{3}+12\,{x}^{2}$$
A: Your formula for the distance is correct. So if you want this distance to be equal to $\sqrt{14} $ then you need to solve 
$$\sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}=\sqrt{14}.$$
A: Let $(x+1)^2=t$, where $t\geq0$.
Thus, we need to solve $t+(t-2)^2=14$.
I hope now it's clear. 
A: Consider : $y': = y - 2;$ $x':= x+1.$
Express the equations in terms of a translated coordinate system $Y'X'$, with its origin at $(-1,2)$. A translation does not alter  length, it suffices to find the distance of the parabola' (x',y') to (0,0).
$\star)$ $ y' = -(x')^2 + 2.$
(Distance)$^2$:
$(x')^2 + (y')^2 =$
$ - y' + 2 + (y')^2 = 14$.
$(y')^2 - y' -12 = 0.$
$(y' -4)(y' +3) = 0.$
$y'_1=4;$ $y'_2=-3.$
No solution for $y'_1= 4.$
$y'_2 = -3$ yields: $ x' = {^+_-}√5$.
And back to the original coordinates $x,y$:
$(√5-1,-1)$ and $(-√5 -1,-1)$.
