Find an equation of the tangent line to the curve at the given point.
2x4 + 4x2 − 7x, (1, -1)
I know i would need to use m = lim f(x)-f(a)
x->a x-a
but then i have lim 2x^4+4x^2-7x-1
x->1 x-1
i am not quite sure how to factor that limit so i can cancel the x-1 and get a equation that i can plug my x=1 value into and get my slope so i can use the formula:
y-h = m(x-a)
The derivative of the function at some point is the slope of the tangent line of the curve at that point. If y = 2x^4 + 4x^2 − 7x, and you want to find the tangent line equation at (1, -1), then the tangent line equation is: y = -1 + y'(1)(x - 1). Using the general derivative rules, we find that y' = (2x^4)' + (4x^2)' − (7x)' = 8x^3 + 8x - 7. So y'(1) = 8(1)^3 + 8(1) - 7 = 8 + 8 - 7 = 9.
Therefore the tangent line at (1, -1) is y = -1 + 9(x - 1)