Original post by subbhy

Hi, could someone help me with this question please?

I worked out the coordinate as (2,27)

Differentiated to get y =10x

Not sure what to do now.

Answer is y = 20x - 13

I worked out the coordinate as (2,27)

Differentiated to get y =10x

Not sure what to do now.

Answer is y = 20x - 13

As its just a parabola, you could sketch the curve/tangent line and that may help you think about out, but the tangent is a line so you need the gradient "m" (at the tangent point) and a point (maybe the tangent point) on the line to get "c" ...

(edited 10 months ago)

Original post by mqb2766

As its just a parabola, you could sketch the curve/tangent line and that may help you think about out, but the tangent is a line so you need the gradient "m" (at the tangent point) and a point (maybe the tangent point) on the line to get "c" ...

sorry, not sure what you mean. Don’t I have to differentiate?

Original post by subbhy

sorry, not sure what you mean. Don’t I have to differentiate?

To get the gradient of the line then yes. What do you understand about a tangent?

(edited 10 months ago)

Original post by subbhy

Hi, could someone help me with this question please?

I worked out the coordinate as (2,27)

Differentiated to get y =10x

Not sure what to do now.

Answer is y = 20x - 13

I worked out the coordinate as (2,27)

Differentiated to get y =10x

Not sure what to do now.

Answer is y = 20x - 13

After differentiating, write dy/dx =10x NOT y=...

The tangent is the line touching the curve at the pt (2, 27). The gradient of this line is equal to the value of dy/dx when x=2... so you can calculate this gradient, m. Then find the line with this gradient passing through (2, 27).

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