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Find the tangent line to the graph of a cubic at a particular point

Consider the cubic curve

Find values for and such that the line is tangent to the graph of at the point .

There is a second line passing through the point which is tangent to the graph of at the point . Find the values of and .

First, we sketch the graph of on the interval :

We compute the derivative of ,

So, the slope of the tangent line at the point is

Then, since is the tangent line to the curve at it must be on the curve at this point, so

Therefore, the line is , i.e., .

Next, if there is another line, say , tangent to at a point we know it has slope given by

Since it must pass through the point (by hypothesis) we must have

So the line is of the form,

Finally, since the point is on both this line and the curve we have,

Since these are both equal to , they must be equal to each other,

Since we then have .

Thus, , and are the requested values.

Find the point at which a given line is tangent to a given curve

Consider the curve given by the cubic equation

Show that the line is tangent to this curve and find the point of tangency. Determine if this line intersects the curve anywhere else.

First, we compute the derivative of the curve,

For the line to be tangent to the curve it must be at a point such that (since this line has slope , the derivative of the curve must be ). So,

Next, for the line to be tangent to the curve the point must actually be on the curve. So, we test out the two possible values of .

If , we have . So is not tangent to the curve at .

If , we have ; hence, is tangent to the curve at .

This tangent line also intersects the curve at .

Find values of constants so two polynomials intersect with the same slope at a point

Let and be the polynomials:

Find values of such that and intersect at the point and have the same tangent line at this point.

Since we want and to intersect at the point we must have . This gives us the equations,

and,

Next, we compute the derivatives so that we can find the slope of the tangent lines at this point,

Since these must be the same at the point we have

Then from above we know ; thus, .

Therefore, the values of the constants are

Find constants so a quadratic has a given tangent at a particular point

Find values for and such that the graph of has a tangent line given by at the point .

First, we compute the derivative

So, if the line is tangent to at the point we must have

Next, the point must be on the graph of , i.e., we must have . Thus,

So, the values are and .

Find the points at which the given function has slope zero

Consider the function

Find the points at which the graph of at has slope zero.

First, the derivative is given by

Then, the requirement that the graph of has slope zero at is asking for the points at which . So, we solve,

Find points of a function at which the tangent line has specified values

Consider the function

Find the points at which the slope of is

1. 0;
2. -1;
3. 5.

The derivative is given by

Then,

1. The slope is 0 means

2. The slope is -1 means

3. The slope is 5 means

Find points at which the tangent to a given function is zero

Find the points at which the tangent line to the function

is horizontal.

First, we compute the derivative,

Setting this equal to zero we have,