Question

How do you find the equation of the line tangent to `f(x)= x^2+3x-7` at x =1?

Answer 1

`y = 5x - 8`

Explanation:

Work out the full co-ordinates of the point at `x = 1` by substituting this value of `x` into `f(x)`.

`f(1) = (1)^2 + 3(1) - 7 = 1 + 3 - 7 = -3`

which gives the point

`(1, -3)`.

Now differentiate the function. Multiply each `x` phrase by the exponent, and then reduce the exponent by `1`.

`f'(x) = 2x^1 + 3x^0 + 0*7x^-1 = 2x + 3`

Find the gradient of the line tangent by inserting your value of `x` into this derivative, to get

`2(1) + 3 = 5`.

Put this into the line equation, `y = mx + c` and then substitute in your values of `x` and `y` from the point `(1, -3)` we found earlier in order to find the value for `c`.

`y = mx + c`
`y = 5x + c`

`- 3 = 5(1) + c`
`- 3 - 5 = - 8 = c`

Put all of this back into the line equation, and you get

`y = 5x - 8`