Question

Using the definition of the derivative find the equation of the line tangent to the curve at x=3 if f(x)=2x^2-13x+5

Answer 1

Take the derivative, what do you get?

Answer 2

4x-13

Answer 3

Precisely. Now evaluate it at x=3 and you will have the slope of your tangent line.

To solve for the equation of the line you'll need:
\[y-y_0=m(x-x_0)\]
where m is the slope we just found and (x_0,y_0) is an arbitrary point on the line. In this case, use x=3 and evaluate f(3). From this you have your point (x_0,y_0). Use this in the above formula for a line with the slope,m, given by f'(3).

Answer 4

Hi Fudger
The general way that you would do this is to find the instantaneous rate of change at that point by using the derivative and also find the coordinates of x=3 by using the original equation. You can then substitute that rate of change and the coordinate into a linear equation (which I forgot the name of) in order to produce an equation with a line going through that coordinate, with the gradient that we found earlier. This happens to be the tangent.

So, find the instantaneous rate of change:
f(x) = 2x^2-13x+5
f'(x) = 4x-13 (derivative)
f'(3) = 4*3-13
f'(3) = -1
Therefore the gradient at that point is -1.

Now find the original coordinate:
f(x) = 2x^2-13x+5
f(3) = 2*(3^2)-13*3+5
f(3) = 18-39+5
f(3) = -16
Therefore the coordinate of the point is (3, -16).

Now use the linear formula to calculate the equation for the tangent:
y-y1 = m(x-x1) (where m is the gradient)
y+16 = -1(x-3)
y+16 = -x+3
y = -x - 13

Therefore the answer to your question is y = -x-13.

You can check this in wolfram alpha: http://goo.gl/5XDDrE
And you can see that the line is a tangent.

Answer 5

So, f(3) would be -16. So you need:
\[(x_0,y_0)=(3,-16)\]

Answer 6

thank you!