At what point along the curve y 3x2 12x 8 is the slope of the tangent line horizontal?

Explanation:

step one: find the derivative of the equation.
#y'=6x^2+6x-12#

Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve.
#y' = 6(x^2+x-2)#
#y' = 6(x+2)(x-1)#
#x= -2, 1#

Step three: plug the x-values found in step 2 back into the original equation to get the y-coordinates of the points on the curve.

#y(-2)= 21#
#y(1)= -6#

Step four: write out the coordinates of the points with a slope of zero.
(-2,21) and (1,-6)

Consider the following line.

graph{y = 0x + 1 [-10, 10, -5, 5]}

What can we say about this line?

We can say that it is horizontal. Let's look at the slope.

#m = (y_2 - y_1)/(x_2 - x_1) = (1 - 1)/(0 - 5) = 0/-5 = 0#

The slope is #0#. So, the slope of any horizontal line is #0#, since any other horizontal line will be parallel to this one.

Hence, we need to find the points on the derivative where the slope of the tangent is #0#.

The derivative can be found using a combination of the sum/difference and power rules.

#y' = 6x^2 + 6x - 12#

The slope of the tangent is given by plugging in a point, #x = a#, into the derivative.

Hence, we can set #y'#, the slope, to #0# and solve for #x#.

#0 = 6x^2 + 6x - 12#

#0 = 6(x^2 + x - 2)#

#0 = (x + 2)(x - 1)#

#x = -2 and 1#

All that is left to do is determine the corresponding y-coordinates that the function passes through.

#y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 1 = 2(-8) + 3(4) + 24 + 1 = 21#

AND

#y = 2(1)^3 + 3(1)^2 - 12(1) + 1 = 2 + 3 - 12 + 1 = -6#

Hence, the points where the tangent is horizontal are #(-2, 21)# and #(1, -6)#.

Hopefully this helps!

2.3 pt 2 14

At what point along the curve y 3x2 12x 8 is the slope of the tangent line horizontal?

Transcribed Image Text:Find the points on the curve y = 2x3 + 3x2 - 12x + 8 where the tangent line is horizontal. (х, у) (smaller x-value) (х, у) (larger x-value) Need Help? Watch It Read It

At what point along the curve y 3x2 12x 8 is the slope of the tangent line horizontal?

At what point along the curve y 3x2 12x 8 is the slope of the tangent line horizontal?

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