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. 
$$
d(x) = \sqrt{(x+1)^2 + \big(-(x+1)^2 + 2\big)^2}
$$

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

