How would i find the tangent to the quadratic 2x^2 -12x + 14 at the point (1,4)? Do we take the derivative of the equation? Thanks

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- Mar 28th 2006, 11:22 PMclassicstringsFind the tangent.
How would i find the tangent to the quadratic 2x^2 -12x + 14 at the point (1,4)? Do we take the derivative of the equation? Thanks

- Mar 29th 2006, 02:05 AMAradesh
yes, that's right, you need to differentiate it.

let

$\displaystyle f(x) = 2x^2-12x+14$

when $\displaystyle x = 1$

$\displaystyle f(x) = 2-12+14 = 4$

so the point (1,4) does lie on the curve.

$\displaystyle f'(x) = 4x-12$

$\displaystyle f'(1) = 4-12 = -8$

so the gradient of the curve at (1,4) is -8.

you could also find the equation of the tangent line that touches the curve at (1,4) but i don't think that's what you need to do. - Mar 29th 2006, 09:02 PMclassicstrings
Yeah the equation is y = -8x + 12 after simple substitution. Thanks!