Question

Find the equation of the line which is tangent to the graph of the function f(x)=3x^2-3x and which is parallel to the x-axis

Answer 1

The curve equation f(x) = 3x² - 3x

y = 3x² - 3x

Differentiating on each side with respect to x.

y' = 6x - 3

 

Now find the tangent point.

Tangent line parallel to the x axis and the slope of the horizontal line is zero.

So equate the y' = 0

6x - 3 = 0

6x = 3

x = 3/6

x = 1/2

Using the x - value find the corresponding y - value with the curve.

y = 3(1/2)² - 3(1/2)

y = 3(1/4) - (3/2)

y = (3/4) - (3/2)

y = ( 3 - 6)/4

y = (- 3)/4

Tangent point is [(1/2), (- 3/4)].

 

To find the tangent line equation, substitute the values of m = 0 and (x , y) = (1/2, - 3/4)  in the slope intercept form of an equation y = mx + b.

- 3/4 = 0(1/2) + b

b = - 3/4

Substitute m = 0 and b = -3/4 in y = mx + b.

y = 0(x) + (- 3/4)

y = - 3/4

Equation of tangent line is y = -3/4.