Question 1149607
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The derivative of the function for the given curve is the slope of the curve at any point.<br>
{{{y = 3+2x-2x^2}}}
{{{dy/dx = 2-2x}}}<br>
The slope of the given line is 1/2; we want the slope of the normal to the curve to also be 1/2.  That means we want the slope of the curve at that point to be -2.<br>
{{{2-2x = -2}}}
{{{4 = 2x}}}
{{{x = 2}}}<br>
{{{y = f(x) = 3+2(2)-2^2 = 3}}}<br>
The slope is 1/2; the point on the curve is (2,3).  The equation of the normal is<br>
{{{y = (1/2)x+2}}}<br>
A graph, showing the given curve (red), the given line (green), and the normal to the curve parallel to the given line (blue).<br>
{{{graph(400,400,-2,4,-1,5,3+2x-x^2,.5x+1.5,.5x+2)}}}<br>