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
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
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