Differential Calculus?

+1 vote
Find the slope of the tangent line to the given curve at the given value of x. Find the equation of each tangent line- y=8-x^2, at x=1
asked Jan 31, 2013 in CALCULUS

+1 vote

Tangent line is y = 8 - x2

Substitute x = 1 in the equation.

y = 8 - (1)2 = 8 - 1 = 7

Point (x, y) = (1, 7)

Slope is y ' = (8 - x2) '

y ' = 0 - 2x

But x = 1, so y ' = -2

There fore slope(m) y' = -2

The "point-slope" form of the equation of a straight line is: (y - y₁) = m(x - x₁)

substitute (x, y) = (1, 7) and m = -2 in the equation.

y - 7 = -2(x - 1)

y - 7 = -2x + 2

y - 7 + 2x = 2

Subtract 2 from each side.

2x + y -9 =0.

The slope of the tangent line is 2x + y -9 =0.

The slope of the tangent line to the implicit curve y = 8 - x2at the point (1, 7) is - 2.

The tangent line equation  is y = - 2x + 9.

The curve is y = 8 - x2  and x = 1.

when, x = 1, y = 8 - (1)2 = 8 - 1 = 7.

So, the point is (1, 7).

Differentiate the curve with respect to ' x '.

y ' = - 2x.

When, x = 1, y ' = - 2(1) = - 2.

y ' = - 2.

This is the slope (m ) of the tangent line to the implicit curve at the point (1, 7).

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Slope (m) = - 2.

Now, the tangent line equation is y = - 2x + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (1, 7).

7 = (- 2)(1) + b

b = 7 + 2

b = 9.

The tangent line equation  is y = - 2x + 9.