Question

How do you find an equation at a tangent line to the curve `y = (5+4x)^2` at point p = (7, 4)? y1=? y2=? ahh pleasee help...i am soo confused here :/?

Answer 1

To find the equation of the tangent line to the curve `y = (5 + 4x)^2` at point `(7,4)` we need to find the gradient/slope of the tangent line at the point `(7,4)` to do this, we need to differentiate our function.

Note that the point does not exist on that curve, so I will use a point that does exist on the curve, which would be `(7,1089)`
I got to that point by simply setting `x=7` in the equation to get `y`

`dy/dx f(x) = dy/dx (5 + 4x)^2`

`f'(x) = 8(5+4x)`

then we sub in the `x` value from the given point.

`f'(7) = 8(5+4(7)) = 264`

now we have got the gradient or `m` at the point where `x=7`
we can start using our formula for a straight line, which is:

`y = mx + c`

and we need to solve for `c` so we use our known point.

`1089 = (264)(7) + c`
`c = - 759`

thus the equation to the line tangent to the curve at the point where `x=7` would be.

`y = 264x - 759`