Need help with tangent lines/equations and derivatives please

[kristoball]

Hello, I just have two examples of problems involving tangent lines and equations.

The first, I'm given the equation of a tangent line to a graph, and the point of the graph it touches, then i'm asked to find the derivative.
Here's the actual question:

If y = -10x+16 is an equation of the tangent line to the graph f at the point (5, -34) then what is f`(5)?

Now I thought that the slope of a tangent line = the derivative of the function, so because -10 is a constant, would the answer by 0 through the constant laws, or is it -10, or am I going about this entirely wrong? I would appreciate any help! Thanks.

The second problem asks me to find the tangent line of an equation, given a point of the original function.
Here's the actual question:

Find the equation of the tangent line to y = x⁴-2x³-7x+8 at the point (-1, 17).

My first guess on what to do would be to find the derivative to get the tangent slope. When I do, I get 4x³-6x²-6, but I don't really know where to go from there. I thought a slope had to be a constant. Once again, any help would be greatly appreciated! Thank you all again =)

 

[pappus]

Hello, I just have two examples of problems involving tangent lines and equations.

The first, I'm given the equation of a tangent line to a graph, and the point of the graph it touches, then i'm asked to find the derivative.
Here's the actual question:

If y = -10x+16 is an equation of the tangent line to the graph f at the point (5, -34) then what is f`(5)?

Now I thought that the slope ... is it -10, <--- that's correct!

The second problem asks me to find the tangent line of an equation, given a point of the original function.
Here's the actual question:

Find the equation of the tangent line to y = x⁴-2x³-7x+8 at the point (-1, 17).

My first guess on what to do would be to find the derivative to get the tangent slope. When I do, I get 4x³-6x²-6 <--- well, that's really a near miss
, but I don't really know where to go from there. I thought a slope had to be a constant. Once again, any help would be greatly appreciated! Thank you all again =)

1. In general your considerations are OK:

\(\displaystyle f(x)=x^4-2x^3-7x+8~\implies~\boxed{f'(x)=4x^3-6x^2-7}\)

2. The slope (or gradient) of f at x = -1 is evaluated by

\(\displaystyle f'(-1)=4(-1)^3-6(-1)^2-7=-17\)

3. Now you know the coordinates of a point of the tangent and the slope of the tangent. Use the point-slope-form of the equation of a line: If the straight line passes through \(\displaystyle P(x_1, y_1)\) and has the slope m then the equation of the line is:

\(\displaystyle (y-y_1)=m(x-x_1)\)

Plug in the values you know to determine the equation of the line.

4. You should come out with y = -17x