Calculating the equations for the tangent/normal lines

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Homework Help Overview

The discussion revolves around calculating the equations for tangent and normal lines based on given slopes and points. Participants are exploring the relationships between tangent slopes and their corresponding normal lines, particularly in cases where slopes are zero or undefined.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to derive equations for tangent and normal lines based on calculated slopes. There are questions regarding the implications of tangent slopes being zero or undefined and how that affects the normal line equations.

Discussion Status

Some participants have provided insights into the relationships between tangent and normal slopes, noting that a tangent slope of zero leads to a vertical normal line, while an undefined tangent slope results in a horizontal normal line. However, there is no explicit consensus on the final equations, and some participants express uncertainty about their calculations.

Contextual Notes

Participants are working with specific points and slopes, and there is mention of given information about points such as (-1, -2). The discussion reflects confusion over how to apply this information correctly in their calculations.

fulton33
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Homework Statement
Use the table to write the equation for tangent lines at given values of x.
Relevant Equations
y-y0=m(x-x0)
IMG_3607.jpeg

9. When I do this problem I know my slope is -3 because f'(2)=-3. I then went and substituted and got
y+5=-3(x-2) which simplified to y=-3x+1

10. I get lost here because the tangent slope would be 0, which would give me the equation y=-2. The normal means perpendicular and the perpendicular slope to 0 is undefined. Not sure if that is right and what to do after.

11. I did the same steps in 9. The slope is 3 and I get the equation y-4=3(x+1) which simplifies to y=3x+7

12. I am lost here as well. The tangent slope would be DNE, which would mean the normal slope to be 0. When I plug 0 in for m and (2,0) for x and y I get y=0. I think that is wrong.
 
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fulton33 said:
10. I get lost here because the tangent slope would be 0, which would give me the equation y=-2. The normal means perpendicular and the perpendicular slope to 0 is undefined. Not sure if that is right and what to do after.
If the slope of the tangent line is 0, its normal will be a vertical line of the form x = k, where k is the x-value at the point of tangency.
fulton33 said:
12. I am lost here as well. The tangent slope would be DNE, which would mean the normal slope to be 0. When I plug 0 in for m and (2,0) for x and y I get y=0. I think that is wrong.
This is the converse of #10. If the tangent slope is undefined, a line perpendicular to it will have slope 0.
 
Mark44 said:
If the slope of the tangent line is 0, its normal will be a vertical line of the form x = k, where k is the x-value at the point of tangency.
This is the converse of #10. If the tangent slope is undefined, a line perpendicular to it will have slope 0.

10. Does that mean for number 10 the equation would be x=0 because at x=-1 f'(-1)=0?

12. Does that mean that for number 12 y=0 would then be correct?
 
fulton33 said:
Does that mean for number 10 the equation would be x=0 because at x=-1 f'(-1)=0?
No. You're given information about the point (-1, -2).
fulton33 said:
Does that mean that for number 12 y=0 would then be correct?
Yes.
 
Mark44 said:
No. You're given information about the point (-1, -2).
Yes.
That makes sense about 10. That would make it x=-1?
 

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