Question 202262
Find the vertex of:
{{{f(x) = x^2-10x+21}}}
The x-coordinate of the vertex is given by:
{{{x = (-b)/2a}}} and in this problem, a = 1, b = -10, so...
{{{x = -(-10)/2(1)}}}
{{{x = 5}}} To find the y-coordinate, substitute this into the given quadratic equation and solve for y (f(x)).
{{{y = x^2-10x+21}}} Substitute x = 5.
{{{y = 5^2-10(5)+21}}} Evaluate.
{{{y = 25-50+21}}}
{{{y = -4}}}
The vertex is at (5, -4)
This is a minimum point on the curve because the positive value of the {{{x^2}}} coefficient indicates that the parabola opens upwards.
{{{graph(400,400,-5,10,-5,5,x^2-10x+21)}}}