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

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Answer 1 · 2015-09-11T12:37:51

$y = -14x+55$

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
enter image source here

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