Question

How do you find the equation of the tangent line to the curve when x has the given value: `f(x) = -6 – x^2` ; x = 7?

Answer 1

`y=-14x+43`

Explanation:

The given function is -

`y=-6-x^2`
At `x=7 ; y= -6-(7^2)=-6-49=-55`

Point `(7, -55)` is common both to the curve and the tangent.

At `x=7` the slope of the curve is the same as slope of the tangent.

`dy/dx = -2x` is the slope of the curve.

At point `x=7` the slope of the curve is -

At `x=7; m=-2(7)=-14`

The tangent passes through the point `7,-55`; its slope is `-14`

The equation of the tangent is -

`mx+c=y`
`(-14)7+c=-55`
`-98+c=-55`
`c=-55+98=43`
`y=-14x+43`

!