#### To determine

**To find: **the point where the given curves have common tangent by using graph.

#### Answer

the point where the given curves have common tangent is (2, 6).

#### Explanation

**1) Concept:**

Sketch the graph of given curves.

From graph find point where graphs touch each other.

We have to find the point where both graphs have common tangent.

For graphing common tangent, we have to find slope of tangent differentiate any of the curve and use slope point form of the line.

**2) Formula:**

i. Slope of tangent

mtanΑ=dydx

ii. Equation of tangent through a, b is

y-b=mtanΑ (x-a)

**3) Given:**

Equation of curves y=x3-3x+4 and y=3x2-x

**4) Calculations:**

Consider y=x3-3x+4

Differentiate with respect to x,

dydx=ddx(x3-3x+4)

By using sum and difference rule,

dydx=ddx(x3)-ddx(3x)+ddx(4)

By using power rule, constant multiple rule and constant function rule,

dydx=3x2-3ddx(x)+0

⇒dydx=3x2-3

Which is slope of tangent line

Now consider y=3x2-x=3x2-3x

Differentiate with respect to x,

dydx=ddx(3x2-3x)

By using difference rule,

dydx=ddx(3x2)-ddx(3x)

By using constant multiple rule,

dydx=3ddx(x2)-3ddx(x)

By using power rule,

dydx=3(2x)-3

⇒dydx=6x-3

Which is slope of tangent line to y=3x2-x at x.Let x be the x-coordinate of point of common tangency, then since the slope of both tangent lines is same,

Therefore,

3x2-3=6x-3

⇒3x2-6x=0

⇒3x(x-2)=0

⇒x=0 and x=2

But x = 0 is not a common point,

At x = 0, slope of both tangents becomes – 3 but curves did not intersect because

at x =0, x3-3x+4=4

at x =0, 3x2-x=0

At x = 2, slope of both tangents becomes 9 and the curves touches at (2, 6)

Graph the equations y=x3-3x+4 and y=3x2-x.

From graph, common point is (2,6) so given curves has common tangent at point (2, 6).

Equation of tangent through 2,6 with slope mtanΑ=9 is,

y-6=9 x-2

y-6=9x-18

y=9x-12

**Conclusion:**

The point where the given curves have common tangent is (2, 6).