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TANGENTS AND NORMALS find the points on the curve y=3x^2 – 9x + 8 at which the tangents are equally inclined with the axes

TANGENTS AND NORMALS
find the points on the curve y=3x^2 – 9x + 8  at which the tangents are equally inclined with the axes

Grade:12

3 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

Slope = \pm 1
y = 3x^2 - 9x + 8
y' = 6x - 9
6x - 9 = \pm 1
6x = 9 \pm 1
x = \frac{4}{3}, \frac{5}{3}
Put in curve:
y = 3(\frac{4}{3})^2 - 9.\frac{4}{3} + 8
y = \frac{16}{3} - 4
y = \frac{4}{3}
(\frac{4}{3}, \frac{4}{3})
Similarly you will get for x = 5/3.
grenade
2061 Points
8 years ago
tangent means slope and slope means dy/dx thus u can diffrentiate both sides and get the answer
Sumer Singh
24 Points
8 years ago
You must add more information to the answer by providing why the slope is =1/-1 and whose slope.
 
Let me add some more values to the answer.
 
Any line inclined equally with axes makes angle of 45 degrees or 135 degrees with positive direction of x-axis.
 
As the slope of tangent is tan45 degree or tan 135 degree which gives slope of tangent=1/-1.
 
Thats all.
 
I hope you liked my explanation.
 
Sumer Singh

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