Answer
a) In standard form, $f(x) = 3(x-1)^2-2$.
b) Graph is attached.
c) $(1, -2)$ is the minimum point.
Work Step by Step
a) To find the standard form of some function $f(x) = ax^2+bx+c$, the standard form would be $f(x) = a(x-h)^2+k$ where $h = -\frac{b}{2a}$ and $k = f(h)$. This is a standard result derived in the book simplifies the algebra and gives a closed form for the end result.
For this problem, $a= 3,b = -6, c = 1$. Plugging above, we get $h =1 , k = f(1) = -2$ and hence $f(x) = 3(x-1)^2-2.$
b) To plot the graph, we plot the vertex and the y-intercept and join them using a smooth curve; we also note that there would be a line of symmetry about a vertical axis passing through this vertex. This allows us to be able to draw the parabola using only two points.
The vertex can be easily deduced from the standard form; it is the point $(h, k)$ so in this case $(1, -2)$.
The y-intercept can be easily deduced from the original form; it is the $c$ (this can be verified by plugging in $x=0$ into $f(x)$ and seeing that it reduces to $c$). Therefore, in this case, the y-intercept is $(0, 1)$.
c) The maximum or minimum of a quadratic function is attained at the vertex. To determine whether the vertex is a maximum or minimum, we look at $a$:
- if $a>0$, this tells us the function will grow towards positive infinity and hence, the vertex is a minimum.
- if $a<0$, this tells us that the function will grow towards negative infinity and hence, the vertex is a maximum.
Looking at $a$ in this case tells us that the vertex $(1, -2)$ is a minimum.
