Question
A graphing celculator is recommended.
(a) The curve with ecuation
is called the Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point ( 1,2 )
(t) At what points does this curve have horizontal tangents? (Order your answers from smallest to largest
, then from smallest to largest
)
(
(b) by graphing the curve and the tangent lines on a common screen.
(a) The curve with ecuation
(t) At what points does this curve have horizontal tangents? (Order your answers from smallest to largest
Ask by Simpson Fuentes. in the United States
Mar 20,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
To find the tangent line at (1,2), the equation is
. The curve has horizontal tangents at
,
, and
.
Solution
It seems that the system does not currently support plotting the graphs directly. However, I can guide you on how to graph the curve and the tangent line using graphing software or a graphing calculator.
Instructions for Graphing
-
Graph the Curve:
- Use the equation
. - You can rearrange it to
to plot both the positive and negative branches of the curve.
- Use the equation
-
Graph the Tangent Line:
- Use the equation
. - This is a straight line with a slope of
and a y-intercept of .
- Use the equation
-
Set the Viewing Window:
- For both graphs, set the x-axis from approximately -4 to 2 and the y-axis from -3 to 3 to capture the relevant features of the curve and the tangent line.
Points to Mark
- Mark the point of tangency (1, 2) on the graph.
- Also, mark the points where the curve has horizontal tangents:
, , and .
Conclusion
By following these instructions, you should be able to visualize the Tschirnhausen cubic and its tangent line effectively. If you have access to graphing software like Desmos, GeoGebra, or a graphing calculator, you can input the equations directly to see the graphs.
Answered by UpStudy AI and reviewed by a Professional Tutor
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The Deep Dive
To find the equation of the tangent line to the curve
at the point
, we first need to differentiate the equation implicitly. By applying implicit differentiation, we find that
. Then, substituting the point
, we have
, which simplifies to
or
. The tangent line has the equation
, or in slope-intercept form,
.
For horizontal tangents, we set
. This requires
, which results in
, giving
and
. To find the corresponding
values, substitute back into the original equation. For
:
→
. For
:
→
. Thus, the points with horizontal tangents are:
,
, and
.
Now, graphing the curve
with the tangent line
will visually demonstrate how the curve behaves at the specified point and where it intersects at horizontal tangents!