[SOLVED] Quadratic Equation

Quote:

Originally Posted by Ilsa

Sketch the curve $y = 2x^2 -4x + 1$ , indicating the coordinates of the turning point and the exact values of the x-intercepts.

I am able to deduce the coordinates of the turning point, but how do we calculate the values of the x-intercepts?

According to my calculation, the coordinates of the turning point are (1, -1).

The x intercepts are simply the roots of the function. So set $2x^2 - 4x + 1 = 0$ and solve for x.

As far as the turning point is concerned, if we have a quadratic function $y = ax^2 + bx + c$ then the line of symmetry is $x = -\frac{b}{2a}$ . The vertex is either the lowest (when a is positive) or the highest (when a is negative) point on the parabola.

-Dan