Question

A curve is such that 3x^2 - 4x + 1. The curve passes through the point (1, 5).

(i) Find the equation of the curve.
(ii) Find the set of values of x for which the gradient of the curve is positive.

Options:
A) (i) y = 3x^2 - 4x + 1; (ii) x > 0
B) (i) y = 3x^2 + 4x - 1; (ii) x < 0
C) (i) y = 3x^2 - 4x - 1; (ii) x > 0
D) (i) y = 3x^2 + 4x + 1; (ii) x < 0

Answer 1

The equation of the curve and the set of values of (x) for which the gradient of the curve is positive is: Option A: y = 3x² - 4x + 1; (x > 0).

What is the equation of the curve?

The equation of the curve is already given as:

y = 3x² - 4x + 1

To find the equation of the curve, we can use the given point (1, 5) to solve for the constant term in the quadratic equation.

Substituting the coordinates of the point into the equation, we get:

5 = 3(1)² - 4(1) + 1

5 = 3 - 4 + 1

5 = 0 + 1

This shows that the constant term is 1, so the equation of the curve is truly y = 3x² - 4x + 1

To find the set of values of (x) for which the gradient of the curve is positive, we need to find the values of (x) for which the derivative of the curve is positive.

The derivative of the curve is:

dy/dx = 6x - 4

At dy/dx = 0

x =2/3

This means that it is positive when (x > 2/3).

Therefore, we conclude that the set of values of (x) for which the gradient of the curve is positive is x > 0