Equation of the tangent line

Geometry Level pending

Find the equation of the line tangent to the curve y = 3 x 2 − 4 x y=3x^2-4x and parallel to the line 2 x − y + 3 = 0 2x-y+3=0 .

y = 2 x + 2 y=2x+2 y = 2 x + 3 y=2x+3 y = 2 x − 3 y=2x-3 y = 2 x − 2 y=2x-2

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1 solution

Harry Jones
Dec 29, 2016

By differentiating the curve equation we get its slope at a point suppose ( x , y ) (x,y) d y d x = 6 x − 4 = 2 \dfrac{dy}{dx}=6x-4=2 ; since slope of line is 2 2 . Hence x = 1 x=1 . Putting value of x x in the curve equation we get y = − 1 y=-1 . By slope point equation of the line is y + 1 x − 1 = 2 \frac{y+1}{x-1}=2 implying y = 2 x − 3 y=2x-3

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did the same !!

Ayon Ghosh - 3 years, 8 months ago

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