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The curve y = ax2 + bx + c passes through the point (2, 32) and is tangent to the line y = 4x at the origin. Find a, b, and c.
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4a + 2b + c = 32 2ax + b = 4 b = 4 c = 0 a = 6
yes thats the right answer but i dont know how to get it....
Let's go step by step: 4a + 2b + c = 32 (The parabola passes through (2,32), so sub x=2, y=32) Differentiating the original equation w.r.t. x to get 2ax+b, equate the slope to the slope of the tangent line y=4x, slope =4, these give 2ax + b = 4, but at (0,0), x=0, so b=4 Since the curve passes through origin, c=0, in ax^2+bx+c=0 Substitute c=0, b=4 into 4a+2b+c=32 to get a=6 => b = 4 c = 0 a = 6
thanks so much
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