# How to do this?

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equations of the normals to the curve

#y=2^x - 7# at the points where #x=1# and #x=-1.# calculate the coordinates of the point of intersection of these normals.

equations of the normals to the curve

##### 1 Answer

The Normals at

# (-5/3-ln2, 4/(3ln2) -9/2)#

#### Explanation:

The gradient of the tangent at any particular point on the function is given by the derivative at that point. The normal is perpendicular to the tangent, so their products will be

Differentiating wrt

When

Let

**At A:**

gradient of tangent

gradient of normal

So normal passes through

# y - (-13/2) = -2/ln2(x-(-1)) #

# y +13/2 = -2/ln2x-2/ln2 #

# y = -2/ln2x-2/ln2 -13/2# ..... [1]

**At B:**

gradient of tangent

gradient of normal

So normal passes through

# y - (-5) = -1/(2ln2)(x-1) #

# y +5 = -1/(2ln2)x+1/(2ln2) #

# y = -1/(2ln2)x+1/(2ln2) -5# ..... [2]

**Point of Intersection**

Eq [1] = Eq[2] we get

# -2/ln2x-2/ln2 -13/2 = -1/(2ln2)x+1/(2ln2) -5 #

Multiply by

Subs into Eq [2]:

We can confirm this graphically: