# Question on slope of a tangent line

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?

• It looks good to me. – José Carlos Santos Jun 1 at 11:13
• Note that the negative value of $x$ is the $x$ coordinate of a point where the $y$ coordinate of the graph and the slope of the graph are equal and positive, unlike the point shown in the figure where the $y$ coordinate is negative. In other words, there are two solutions to the words describing the problem, and you found both of them, but only one matches the picture. – David K Jun 1 at 11:37