# Thread: equation of the tangent line to the curve at the given point

1. ## equation of the tangent line to the curve at the given point

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)

2. Originally Posted by rhcp1231
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 gradient is $m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$

in this problem, $m = f'(1) = 8 + 8 - 7 =9$

3. how did you get
?

4. from $f'(x) = 8x^3 + 8x - 7$

5. 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)

6. oooo so the dervitive of the original equation is the slope of the tangent?

7. Originally Posted by rhcp1231
oooo so the dervitive of the original equation is the slope of the tangent?
Yes.
That's what a derivative of a function is.