2 UNIT MATHS QUESTION (1 Viewer)

[ellie95]

hi everyone

I have to find the equation of the tangent to the curve y= x squared + 3x+2 at the point 6

I first differentiated it.Then I substituted 1 into the x spot for the gradient formula. Then i used y-y1 formula. Is that right ??? Please Help????????

 

[Shadowdude]

6 isn't a point. Do you mean... x = 6?

EDIT: For the point (1,6)




So if it's x = 1:



Now you use that y-y1 formula:



m = 5 from what we've figured out, x1 is 1, y1 is 6.

So then:



Rearrange:




or something like that

 

[ellie95]

yeah the points (1,6)

 

[SpiralFlex]

We have the x coordinate of x = 1. To find the y coordinate we can substitute it into the equation.






Now our point of concern which lines on the parabolic curve is



At

The gradient of the tangent is going to be (when you substitute x =6 into y')








So the gradient is 15 at that point. We have a gradient, we have a point, we can now the form:

 

[Shadowdude]

See above on my original post.

 

[ellie95]

thanks so much for your help. i appreciate it

 

[Carrotsticks]

This thread should not be in Site Help. It should be in the Mathematics section.

Can somebody please move the thread?

 

[Sanjeet]

6 isn't a point. Do you mean... x = 6?

EDIT: For the point (1,6)




So if it's x = 1:



Now you use that y-y1 formula:



m = 5 from what we've figured out, x1 is 1, y1 is 6.

So then:



Rearrange:




or something like that

it's 5x+11 not 5x+1 lol

 

[AAEldar]

it's 5x+11 not 5x+1 lol

Nope... definitely is 1.

 

[Sanjeet]

Nope... definitely is 1.

Oh shit, my bad