Three Forms of Quadratic Functions

The y-intercept is the value of $f(x)$ when $x=0$.

In $f(x)=a{x}^2+bx+c$, substituting $x=0$ gives $f(0)=a(0{)}^2+b(0)+c=c$. The y-intercept is simply $c$, readable directly from the equation.

In vertex form $f(x)=a(x-h{)}^2+k$, we must substitute $x=0$ and expand: $f(0)=a(0-h{)}^2+k=a{h}^2+k$. This requires calculation.

In factored form $f(x)=a(x-p)(x-q)$, substituting $x=0$ gives $f(0)=a(-p)(-q)=apq$. This also requires calculation.

Only standard form reveals the y-intercept immediately as the constant term $c$.

In $f(x)=a(x-h{)}^2+k$, the constant $k$ is the y-coordinate of the vertex, NOT the y-intercept. You must compute $a{h}^2+k$ to find the y-intercept, so this form does not give it immediately.

In $f(x)=a(x-p)(x-q)$, $p$ and $q$ are x-intercepts, not the y-intercept. You must multiply $a(-p)(-q)=apq$ to get the y-intercept, so this form does not give it immediately.

Since vertex and factored forms require calculation, the answer cannot be 'all three forms'.

By elimination, only standard form immediately reveals the y-intercept as the constant $c$.

The equation $h(t)=-3(t-2{)}^2+12$ is in vertex form $f(x)=a(x-h{)}^2+k$.

Comparing to $a(t-h{)}^2+k$, we have $a=-3$, $h=2$, and $k=12$. The vertex is at $(2,12)$.

Since $a=-3<0$, the parabola opens downward, so the vertex is a maximum point.

The maximum height is $k=12$ m, occurring at $t=h=2$ s.

In vertex form $a(t-h{)}^2+k$, the vertex x-coordinate is $+h$, not $-h$. Since the equation shows $(t-2{)}^2$, the time is $t=+2$, not $-2$.

The vertex coordinates are $(h,k)$, not $(k,h)$. The vertex is $(2,12)$, not $(12,2)$.

The value 3 is the magnitude of $a$, which controls the shape, not the maximum height. The maximum height is $k=12$.

The vertex is at $(2,12)$, so the maximum height is 12 m at $t=2$ s.