Take the curve y = 2x² - 4 and point P (2,4). a) First, find the straight line tangent to the curve at point P. This will have the gradient of the curve at point P. To find it, differentiate: dy/dx = 4x. At P, this will equal 4(2) = 8. As the tangent has a gradient of 8 and passes through P, its equation can be found using dy/dx = (y₂ - y₁)/(x₂ - x₁), where (x₂ ,y₂) are the coordinates of P and (x₁ ,y₁) are the coordinates of the y-intercept. We know x₁ = 0 since the y-intercept is on the y-axis. We need to find y₁, so: 8 = (4 - y₁)/(2 - 0); 8 * 2 = 4 - y₁; 4 - 16 = y₁; y₁ = -12. The equation of the tangent is y = 8x - 12. b) To find the normal, we need to find its gradient and y-intercept. The gradients of perpendicular lines are inverse. Therefore, the gradient of the normal is 1/8. We use the same method as previously to find the y-intercept: 1/8 = (4 - y₁)/(2 - 0); 1/8 * 2 = (4 - y₁); 4 - 1/4 = y₁; y₁ = 3.75. So, the equation of the normal to the curve at point P is y = 1/8x + 3.75.

5Sin[x]-4=2Cos[2x]

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