Original post by subbhy
Hi, could someone please help me with this question?
We can only do so much under TSR policy, but I will see what I can do. What's the question?
Original post by MindMax2000
We can only do so much under TSR policy, but I will see what I can do. What's the question?
Thanks [attach]1080180[/attach
[quote="subbhy;98720645"]Thanks 
I’d find the gradient of the curve by doing dy/dx
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
I’d find the gradient of the curve by doing dy/dx
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
[quote="subbhy;98720645"]Thanks
@harlz_chalamet has given you the precise instructions, although I think he might have violated TSR policy.
@harlz_chalamet has given you the precise instructions, although I think he might have violated TSR policy.
Original post by harlz_chalamet
I’d find the gradient of the curve by doing dy/dx
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
Please don't include so much detail in your post - the OP should be doing the work here based on minimal hints
Original post by MindMax2000
@harlz_chalamet has given you the precise instructions, although I think he might have violated TSR policy.
They haven't given the full answer, but it's a bit borderline - I would have stuck to 1 or 2 hints and then left the OP to do some work as per TSR guidelines.
Original post by harlz_chalamet
I’d find the gradient of the curve by doing dy/dx
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
then I’d sub in x=2 into dy/dx
The tangent will have the same gradient
sub in x=2 into y=5x^2+7 to get the y-coordinate of the tangent. We now have the exact coordinate for where the tangent touches the curve.
then, to find the equation of the tangent, use:
y-y1=m(x-x1)
sub in the the gradient (m) and the coordinates into the equation and rearrange
Thank you!
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