The figure shows the graph $f(x)=-2x^2+4x+6$ along with the tangent line at the point $(x,y)$
The given point has a nice property: the slope fo the tangent line is equal to the y-value at that point. Find the x-coordinate.
So the slope of the tangent line is the derivative which in this case would be $-4x+4$
This derivative is equal to the y-value at that point. So i will set the two equations equal to each other. .
$-2x^2+4x+6=-4x+4$
$-2x^2+8x+2=0$
$-2(x^2-4x-1)=0$
$x=2\pm\sqrt{5}$ Reject the negative so it is $2+\sqrt{5}$
Is this correct?