Question

Please i need help for this calculus

Answer 1

Step-by-step 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,

$5 = 2 {x}^{2} + 7x - 4$

$2 {x}^{2} + 7x - 9$

$2 {x}^{2} - 2x + 9x - 9$

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

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

$x = 1$

$2x + 9 = 0$

$x = - \frac{9}{2}$

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

$4( \frac{ - 9}{2} ) + 7 = - 11$

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.