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Differential Calculus> TANGENTS AND NORMALS find the points on t...

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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

taniska , 10 Years ago

Grade 12

3 Answers

Jitender Singh

Last Activity: 10 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

Last Activity: 10 Years ago

tangent means slope and slope means dy/dx thus u can diffrentiate both sides and get the answer

Sumer Singh

Last Activity: 9 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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