How do you find the tangent line for this equation and then find its slope given #y=-2x^5 - 7x^3 + 8x^2# at x=1?

Answer 1

Slope of tangent is #-31# and equation of tangent is #31x+y=30#

As #y=-2x^5-7x^3+8x^2#,
at #x=1#, we have #y=-2-7+8=-1#
Hence, we are seeking tangent at point #(1,-1)# on the curve.

We do not first find the equation of the tangent and then its slope. In fact, we find its slope first and then using point slope form of equation, we find the equation of tangent.

Nowslope of tangent is given by the value of its first derivative at that point.

As first derivative of #y=-2x^5-7x^3+8x^2# is #-10x^4-21x^2#,
the slope of tangent is #-10*1^4-21*1^2=-10-21=-31#
and equation of tangent is #(y-(-1))=-31(x-1)#
or #y+1=-31x+31# i.e. #31x+y=30#
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Answer 2

To find the tangent line and its slope, we need to find the derivative of the given equation. The derivative of y = -2x^5 - 7x^3 + 8x^2 is dy/dx = -10x^4 - 21x^2 + 16x.

To find the slope of the tangent line at x = 1, we substitute x = 1 into the derivative equation: dy/dx = -10(1)^4 - 21(1)^2 + 16(1) = -10 - 21 + 16 = -15.

Therefore, the slope of the tangent line at x = 1 is -15.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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