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

• Feb 9th 2010, 06:04 PM
rhcp1231
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)
• Feb 9th 2010, 06:11 PM
dedust
Quote:

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, $m = f'(1) = 8 + 8 - 7 =9$
• Feb 9th 2010, 06:17 PM
rhcp1231
• Feb 9th 2010, 06:19 PM
dedust
from $f'(x) = 8x^3 + 8x - 7$
• Feb 9th 2010, 06:23 PM
blackcompe
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)
• Feb 9th 2010, 06:31 PM
rhcp1231
oooo so the dervitive of the original equation is the slope of the tangent?
• Feb 9th 2010, 06:43 PM
dMh
Quote:

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.