Find the standard (slope-intercept form) equation of the tangent line to the following

Question

f(x)=x^2-3x+1 at x=1

This question is the only question that i can't answer its very confusing can someone please help me thank you

Answer 1

Given: f(x)=x²-3x+1 at x=1

f(1) = 1² -3(1) + 1 = -1

This means the point of tangency is at (1,-1)

We know that the slope-intercept form of a linear equation is y= mx + b, first find the slope using derivatives:

f'(x) = 2x - 3

f'(1) = 2(1) - 3 = -1

Therefore the slope (m) is -1

m = -1

Using the point of tangency and the slope, we can have the linear equation in point slope form

y - y₁ = m (x - x₁)

where: m = -1 and (x₁,y₁) = (1,-1)

y - (-1) = -1(x - 1)

y + 1 = -x + 1

Subtract 1 on both sides of the equation, we'll have the slope intercept form of the equation of the tangent line:

y = -x

Answer 2

f(x) = x^2 -3x +1

f'(x) = 2x-3

f'(1) = 2(1)-3 = 2-3 = -1 = the slope of the tangent line when x=1 and f(1)=y= (1)^2 -3(1)+1 = 1-3+1 = -1

when x=1 y=-1.

f'(1) = -1 at the point (1,-1)

y =(-1)x + b =-x+b plug in the point (1,-1) to solve for b

-1 = -(1) + b

b= 0

y=-x is the equation of the tangent line when x=1