Question

find the equation of the tangent to the curve at given point.

2x³+3x²-4 at (1,1)

Note: (please solve step by step)
no fake solutions are allowed ​

Answer 1

Step-by-step explanation:

step one: find the derivative of the equation.

y'=6x2+6x−12

Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve.

y'=6(x2+x−2)y'=6(x+2)(x−1)x=−2,1

Step three: plug the x-values found in step 2 back into the original equation to get the y-coordinates of the points on the curve.

y(−2)=21y(1)=−6

Step four: write out the coordinates of the points with a slope of zero.

(-2,21) and (1,-6)

here is an example for you

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Answer 2

Step-by-step explanation:

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