Find an equation of the tangent line to the curve 𝑦 = 2𝑥 2 + 3 that is parallel to the line 8𝑥 – 𝑦 + 3 = 0
"8x-y +3=0 --> y = 8x +3 = mx + b"
A line parallel to y = 8x+3 will have the same slope, m=8. The slope is also the derivative of y=2x2+3
y' = 4x = 8 --> x = 2
y = 2(2)^2+3 = 11. So the intersection point is (2,11)
To get the equation for the line going through that point
y-11 = 8(x-2) --> y = 8x-16+11 = 8x-5
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