Simple Calculus problem...?

Question

given: f(x)=(4x^2)-(5x)-2

...so obviously f'(x)=8x-5

the question is: write an equation of the line tangent to f(x) at x=2.

i know it's probably really easy, i'm just having a mental block because i'm tired...

Answer 1

Okay you have the derivative.  Now you can find the slope of the tangent at x = 2 which is f'(2):

f'(2) = 8(2) - 5 = 16 - 5 = 11

Therefore, the slope is m = 11.

Can we find the equation now?  Nope not yet.  We still need a point.  We know that the tangent intersects f(x) at x = 2.  Therefore, we can find that point which will be (2, f(2)) = (2, 4).

Now use the point slope form and m = 11 and (x₁, y₁) = (2, 4):

y - y₁ = m(x - x₁)

y - 4 = 11(x - 2)

y - 4 = 11x - 22

y = 11x - 18

Hope this helps!

Answer 2

the first derivative of the function gives you the slope of the tangent line at a point

you have already determined that the derivative is 8x-5

when x=2, the value of the derivative, hence the slope of the line, is 11

remember now that a straight line can be written in the form:

y=mx+b  where m is the slope and b is the y intercept (the point where the line crosses the y axis)

we already know that m=11

now we have to find out where the tangent line crosses the y axis. to do this, we need to know the y value of the tangent line at x=2

for this, we go back to the original function

f(x)=4x^2 - 5x -2

when x is 2, this function has the value 4

in other words, we know that one point on the tangent line is the point (2,4)

we also know the slope of the line is 11

so when x=0, the value of y is -18

and the equation of the tangent line is

y=11x - 18

(check, when x=2, y=4 as must be the case if the tangent line touches the curve at x=2)

Answer 3

let y = f(x) = 4x² - 5x -2

  

    y' = 8x -5

Slope of the tangent at x = 2 is y' at x = 2

y' = 8(2) -5

y' = m = 11

When x = 2, then y = 4(2)² - 5(2) -2

                           y = 16 -10 -2

                           y = 4

So the point of tangency is (2,4)

now the equation of tangent

we know the slope and one point

y - 4 = 11(x-2)

y -4 = 11x -22

   11x - y + 18 = 0