- Thread starter Islandguy
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Yes, so how can we find the slope of the line tangent to the given function?slope of horizontal line is 0

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Yes:I found the derivative which is -18×-8 but I am stuck here

\(\displaystyle y'=-18x-8\)

This tells us that for a given value of \(x\), what the slope of the line tangent to the curve at \((x,y)\) will be. We want this slope to be zero, because the tangent line is to be horizontal. So, equate the derivative to zero, and solve for \(x\)...what do you get?

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Hello Islandguy. That y is not the same as the y in your exercise, so I would pick a different symbol when writing the slope-intercept form for an arbitrary tangent line to the curve in this exercise.y=mx+b

Y = mx + b

But, we're not asked to report an equation for the horizontal tangent line, so actually we don't need the slope-intercept form at all. Let's just go with m=slope.

As we move along the given parabola in this exercise, x changes. As x changes, so does m. (There are lots of different tangent lines along the parabola, and each has its own slope.)

Did your calculus class mention the following?

The first derivative is a slope. That is, at each point on the curve of a function y, the value of m is y

In other words, the first derivative of a function gives us the tangent line's slope (at each x in the domain). In this regard, the first derivative is a function of x itself: a function of slopes.

When we're interested in where a tangent-line slope is zero, then we're interested in where the first derivative is zero.

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