determine the point at which the slope of the tangent to each parabola is zero

Question

here the the equation of the parabola: y=6x^2 - 3x +4

Answer 1

Hi Shreya,

             The slope of the tangent line is 12x-3 i.e. the derivative of y with respect to x. If we set this equal to 0 we have x=1/4. If you now put this into the equation for the parabola I get y=29/8. 

         So the point is (1/4,29/8)

 

Regards

Jim

Answer 2

Determine the point at which the slope of the tangent
to the parabola y = 6x^2 - 3x + 4 is zero.

1. Using Calculus: Take the derivative which will give a function of x that can be used to find the slope of the parabola at any x.

y = 6x^2 - 3x + 4

y’ = 12x - 3 = 12(x - 1/4)

y’(1/4) = 0. So the tangent line at x = 1/4 will have zero slope.

y(1/4) = (6(1/4) - 3)(1/4) + 4 = (3/2 - 6/2)(1/4) + 4 = -3/8 + 32/8 = 29/8

So the point on the parabola where its tangent line has zero slope is (1/4,29/8)

2. Using Algebra II:

y = 6x^2 - 3x + 4

The tangent line will be horizontal at the vertex.

h = -b/(2a) = -(-3)/(2*6) = 1/4

k = c - ah^2 = 4 - 6(1/4)^2 = 4 - 6/16 = 32/8 - 3/8 = 29/8

The vertex is (1/4,29/8).