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How do you find the equation of a line tangent to #y=-5x^2+4x-3# at x=1?

Answer 1

Aurora Alvarado

The equation of the line is #y(x) = -6x+2#

The equation of the line tangent to the curve #y=f(x)# for #x=barx # is:

#y(x) = f(bar x) + f'(bar x)(x- barx)#

Here:

#f(x) =-5x^2+4x-3#
#f'(x) =-10x+4#

and for #bar x =1#

#f(barx) = -4#
#f'(barx) = -6#

so the t the tangent line is:

#y(x) = -4-6(x-1)#

or:

#y(x) = -6x+2#

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Answer 2

Christopher Allers

To find the equation of a line tangent to a curve at a specific point, you need to find the derivative of the curve and evaluate it at the given point.

The derivative of the curve y = -5x^2 + 4x - 3 is given by dy/dx = -10x + 4.

To find the slope of the tangent line at x = 1, substitute x = 1 into the derivative: dy/dx = -10(1) + 4 = -6.

The slope of the tangent line is -6.

To find the equation of the tangent line, use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the given point on the curve.

Substituting x1 = 1, y1 = -5(1)^2 + 4(1) - 3 = -4 into the equation, we have: y - (-4) = -6(x - 1).

Simplifying the equation gives the equation of the tangent line: y = -6x + 2.

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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.