Question

1. Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function		Line
f(x) = 2x2		2x − y + 2 = 0

y =

2.Find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results.

f(x) = 2(2 − x)²,      (6, 32)

f '(6) =

Answer 1

Given that the function yD2 we have to find an equation of line which is Langent to the curve y=22 and papallel to the line 22-y +220 let at(24340) the line is tangent of yo222 and parallel to 2x4+230 de @ NOCO dy =42 2x-y +220 doc e dy dy - Ak, 24%) . The equation of paralel line at (21 , I). 7-4-2 (2-x) and the equation of tangent line of y=2312 is y-1, 2424 (2-24) The and are identical comparing these 2=474 → 20 L also the point (2,4) lie on ya2x² Y=2302 OT, 4122-{ = { Thus at (22) the line is tangent of 2x=4+2.00 Hence the required equation of the line becomes y-2/2 = 2(2-2) ar y-2=2x-1 or year- 1 + 1 / 2 or, y=2x-2 Given that the function f(x)= 2(2-2)2 Therefore S'(x) = 22 (2-x) (-1) or, f'() = -4 (2-x) = 4(x-2) f'(6)=4-(6-2)= 16 Hence the sbpe ab x76 is