Question 211893
For the quadratic equation, {{{f(x) = 2x^2-12x+18}}} find:
a. The vertex.
The x-coordinate of the vertex can be found by:
{{{x = (-b)/2a}}} where a = 2 and b = -12.
{{{x = (-(-12))/2(2)}}} Evaluate:
{{{x = 3}}} Now substitute this into the given equation and solve for y (f(x)).
{{{y = 2(3)^2-12(3)+18}}}
{{{y = 18-36+18}}}
{{{y = 0}}}
The ordered pair of the vertex is: (3, 0)
b. The equation of the line of symmetry is:
{{{x = 3}}}
cand d. The maximum/minimum value of f(x) is found at the vertex (3, 0),
so f(3) = 0 and, because the coefficient of the {{{x^2}}} term is positive (2), the parabola opens upward so this is a minimum.
{{{graph(400,400,-5,5,-5,5,2x^2-12x+18)}}}