please anyone help me out

Question

Consider the following.
Function Point
f(x) = 5x^2 − 4, (3, 41)
Find an equation of the tangent line to the function at the given point.
y =

Find the function values and the tangent line values at
f(x + Δx) and y(x + Δx) for Δx = −0.01and 0.01.
f(3 + 0.01) =
y(3 + 0.01) =
f(3 − 0.01) =
y(3 − 0.01) =

Answer 1

the tangent line has slope f'(x)
f'(x) = 10x
f'(3) = 30
So, now we have a point and a slope. The line is thus

y-41 = 30(x-3)

Same for the others; just plug in different numbers.