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,
[tex]5 = 2 {x}^{2} + 7x - 4[/tex]
[tex]2 {x}^{2} + 7x - 9[/tex]
[tex]2 {x}^{2} - 2x + 9x - 9[/tex]
[tex]2x(x - 1) + 9(x - 1)[/tex]
[tex](2x + 9)(x - 1) = 0[/tex]
[tex]x = 1[/tex]
[tex]2x + 9 = 0[/tex]
[tex]x = - \frac{9}{2} [/tex]
Now, to find the gradient of the curve let take the derivative of both sides
[tex]5 = 2 {x}^{2} + 7x - 4[/tex]
[tex]0 = 4x + 7[/tex]
[tex]4x + 7[/tex]
Plug in -9/2, let call that point P
[tex]4( \frac{ - 9}{2} ) + 7 = - 11[/tex]
Plug in 1, let call that point Q
[tex]4(1) + 7 = 11[/tex]
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.
