Answer:
The quadratic has a minimum value.
The minimum value is at (3, 2).
Step-by-step explanation:
We are given the quadratic function:
[tex]f(x)=3x^2-18x+29[/tex]
First, since the leading coefficient is positive, this quadratic function will be concave up.
Hence, we will have a minimum value.
The minimum or maximum value is the vertex of the quadratic. The vertex is given by:
[tex]\displaystyle \Big(-\frac{b}{2a},f\Big(-\frac{b}{2a}\Big)\Big)[/tex]
In this case, a = 3, b = -18, and c = 29. Thus, the x-coordinate of the vertex is:
[tex]\displaystyle x=-\frac{-18}{2(3)}=\frac{18}{6}=3[/tex]
And the minimum value is:
[tex]f(3)=3(3)^2-18(3)+29=2[/tex]
