# Given y=-5x^2+2x+(0), find the equation of the tangent line that is horizontal to the curve. The slope-intercept form of the equation of the tangent line that is horizontal to the curve is y=

*print*Print*list*Cite

Given the function `y=-5x^2+2x` :

The line tangent to the graph of the given function will be horizontal at a local maximum, local minimum, or possibly an inflection point. Since the function is a quadratic, the graph is a parabola. Since the leading coefficient is negative, the parabola opens down and the tangent line will be horizontal at the absolute maximum for the function.

Extrema occur only at critical points where the first derivative is zero or fails to exist.

`f(x)=-5x^2+2x`

`f'(x)=-10x+2`

`f'(x)=0 ==>x=1/5`

Since the slope is 0, we know the line is of the form y=k for some constant k. `f(1/5)=-5(1/5)^2+2(1/5)=1/5`

-----------------------------------------------------------

The equation of the tangent line is `y=1/5`

-----------------------------------------------------------