What is the equation of the tangent drawn for f(x)=x-x^2 which is parallel to the line 4x-y-3=0 pl help me °~°

y = 4x + 9/4

Step-by-step explanation:

4x − y − 3 = 0

y = 4x − 3

The slope of the line is 4.

The slope of the tangent line is:

f(x) = x − x²

f'(x) = 1 − 2x

Setting the slope equal to 4:

4 = 1 − 2x

2x = -3

x = -3/2

The point on f(x) is:

f(-3/2) = -3/2 − (-3/2)²

f(-3/2) = -3/2 − 9/4

f(-3/2) = -15/4

Using point-slope form:

y − (-15/4) = 4 (x − (-3/2))

y + 15/4 = 4 (x + 3/2)

y + 15/4 = 4x + 6

y = 4x + 9/4

answer: 298143

step-by-step explanation:

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