Question

Please i need help for this calculus

Answer 1

Explanation:

First thing first, let find the x value of where P and Q both meet y=5, we know that y=5, and y=2x^2+7x-4, so using transitive law,

`5 = 2 {x}^(2) + 7x - 4`

`2 {x}^(2) + 7x - 9`

`2 {x}^(2) - 2x + 9x - 9`

`2x(x - 1) + 9(x - 1)`

`(2x + 9)(x - 1) = 0`

`x = 1`

`2x + 9 = 0`

`x = - (9)/(2)`

Now, to find the gradient of the curve let take the derivative of both sides

`5 = 2 {x}^(2) + 7x - 4`

`0 = 4x + 7`

`4x + 7`

Plug in -9/2, let call that point P

`4( ( - 9)/(2) ) + 7 = - 11`

Plug in 1, let call that point Q

`4(1) + 7 = 11`

So the gradient of the curve at point P (-9/2,5) is -11

The gradient of the curve at point Q (1,5) is 11.