Question

How do you find the equation of the line tangent to the graph of `f(x) = 6 - x^2` at x = 7?

Answer 1

`y = -14x+55`

Explanation:

Given -
`y = 6-x^2`
It is a quadratic function.
Since co-efficient of `x^2` is negative, it is an downward facing 'U'shaped curve.

The slope of the curve at any given point is its 1st erivative.

`dy/dx=-2x`

At`x = 7`; the slope of the curve is `m = -2(7) = -14`

At `x=7`; the y-co-ordinate of the curve is -

`y=6-(7^2)=6-49= -43`

`(7, -43)` is a point on the curve and on the tangent.
The slope of the tangent is -14

Equation of the tangent is
`mx +c=y`
`(-14).7+c=-43`
`-98+c= -43`
`c = -43 +98 = 55`

`y = -14x+55`

Refer the graph

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