Question

Coordinates, gradient and tangent work (see image).

Answer 1

Answer:

a) $P(x,2x^2-5),\ Q(x+h,2(x+h)^2-5)$

b) $4x+2h$

c) $4x$

Step-by-step explanation:

Given the curve

$y=2x^2-5$

a) If the x-coordinate of P is $x$, then the y-coordinate is $2x^2-5,$ so point P has coordinates $(x,2x^2-5)$

If the x-coordinate of Q is $x+h$, then the y-coordinate is $2(x+h)^2-5$ so point Q has coordinates $(x+h,2(x+h)^2-5)$

b) The gradient of the secant RQ is

$\dfrac{y_Q-y_P}{x_Q-x_P}\\ \\=\dfrac{(2(x+h)^2-5)-(2x^2-5)}{(x+h)-x}\\ \\=\dfrac{2(x+h)^2-5-x^2+5}{x+h-x}\\ \\=\dfrac{2(x+h)^2-2x^2}{h}\\ \\=\dfrac{2x^2+4xh+2h^2-2x^2}{h}\\ \\=\dfrac{4xh+2h^2}{h}\\ \\=4x+2h$

c) If $h\rightarrow 0,$ then the gradient $4x+2h\rightarrow 4x$