find the equation of the normal to the graph of the function f(x)=-x^2+8x+4 at the point where x=-1

Question

find the equation of the normal to the graph of the function.

 

Answer 1

Given function f(x) = -x^2+8x+4

y = -x^2+8x+4

At the point x = -1

y = -(-1)^2+8(-1)+4 = -5

The equation is y = -5.

Graph

 

Comments

The normal line equation  is y = (- 1/10)x - (51/10).

Answer 2

The function f (x) = y =  - x² + 8x + 4.

When, x = - 1, y = - (- 1)² + 8(- 1) + 4 = - 1 - 8 + 4 = - 9 + 4 = - 5.

Therefore, the point is (- 1, - 5).

f ' (x) = - 2x + 8.

When, x = - 1, f ' (x) = - 2(-1) + 8 = 2 + 8 = 10.

f ' (x) = 10.

This is the slope (m ) of the tangent line to the implicit curve at (- 1, - 5).

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (- 1, - 5) is - 1/10.

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

Slope (m) = - 1/10.

Now, the normal line equation is y = (- 1/10)x + b.

Find the y - intercept by substituting the the point in the  line equation say (x, y) = (- 1, - 5).

- 5 = (- 1/10)(- 1) + b

b = - 5 - 1/10

b = (- 50 - 1)/10

⇒ b = - 51/10.

The normal line equation  is y = (- 1/10)x - (51/10).