maths help :( find equation of tangent?

Question

the question reads

find the equation of the tangent to the parabola

y= 5 - x^2 at P(2,1)

Thanks for any help, i really have no idea, do i substitute x = 2 into the equation? so y=1?

then what :s sub in y so it equals 4- x^2 meaning x= +-2? please help

Answer 1

f ( x ) = 5 - x ²
f `( x ) = - 2 x
f `( 2 ) = - 4 = m (gradient at x = 2)
Tangent passes thro` (2 , 1)
Equation of tangent line is:-
y - 1 = (- 4) (x - 2)
y - 1 = - 4 x + 8
y = - 4 x + 9

Answer 2

Differentiate the curve to get gradient at any point.
dy by dx = -2x
So, the gradient of the tangent is -2*(x-value)=-4
so the equation of the tangent so far is y=-4x+c
it crosses point 2,1 so when x=2, y=1
so 1=-4*2+c
so c=1--4*2=+9
Therefore the equation of the tangent is y=-4x+9

Answer 3

First, you take the derivative of y = 5 - x^2. This gives you a general equation for the tangent line at any point.

So, y' = -2x

Now, you read off the slope of the tangent line which is -2

Use now the point slope intercept form of a line to write the equation of the tangent line through the point (2,1)
y -y1 = m(x-x1) and there you have it.

Answer 4

you have to find the derivative of the function and the plug the point into it