Question

How do you find the equation of the line tangent to `y = (5+4x)^2` at P = (7, 4)?

Answer 1

y = 264x - 1844

Explanation:

To find the equation of the tangent in the form y = mx + c , where m represents the gradient and c , the y-intercept.

Find value of m by evaluating the derivative at x = 7. Using (7,4) will allow value of c to be calculated.

`y = (5+4x)^2`

differentiate using the`color(blue)" chain rule "`

`dy/dx = 2(5+4x) d/dx(5+4x) = 2(5+4x).4 = 8(5+4x)`

x = 7 : `dy/dx = 8(5+28) = 264 = m" of tangent "`

equation is y = 264x + c

using (7,4) : 4 = 264(7) + c `rArr c = - 1844`

equation of tangent is : y = 264x - 1844

Answer 2

The point `(7,4)` is not on the curve, so there is no tangent line at that point. There are two tangent lines to the curve that pass through the point `(7,4)`

Explanation:

For the function `f(x)=(5+4x)^2`

The point with `x` coordinate `a`, has `y` coordinate `(5+4a)^2`

The slope of the line tangent to the graph of `f` at `(a,(5+4a)^2)` has slope given by `f'(a) = 8(5+4a)`.

Therefore, the equation of the line tangent to the graph of `f` at `(a,(5+4a)^2)` is

`y-(5+4a)^2 = 8(5+4a)(x-a)`.

This can be rewritten as

`y-(25+40a+16a^2) = (40+32a)(x-a)`.

So, we get,
`y=(32a+40)x -16a^2+25`.

Making sure that the line passes through the point `(7,4)` reqyires us to solve the quadratic equation:

`4=(32a+40)7 -16a^2+25` for `a`.

A bit of algebra leads to

`16a^2-224a-301=0`.

The quadratic formula gives us the solutions `a=7+-sqrt1085/4`.

To finish answering the question by finding the equations of the tangent lines, substitute these vales for `a` in

`y=(32a+40)x -16a^2+25`.