Notes
Forms and key features
- Standard form $y = ax^2 + bx + c$: axis $\displaystyle x = -\frac{b}{2a}$, $y$-intercept $c$. Opens up if $a > 0$, down if $a < 0$.
- Vertex form $y = a(x - h)^2 + k$: vertex $(h, k)$. To convert, find $\displaystyle h = -\frac{b}{2a}$ and $k = f(h)$.
- The vertex is the minimum (opens up) or maximum (opens down).
Transformations
- From $y = x^2$: $h$ shifts horizontally (opposite sign), $k$ shifts vertically, $a < 0$ reflects, $|a| > 1$ narrows, $|a| < 1$ widens.
Domain, range and models
- Domain is all real numbers. Range is $y \geq k$ or $y \leq k$.
- Zeros are the $x$-intercepts.
- In $h(t) = -16t^2 + v_0 t + h_0$, the peak is at the vertex and $h_0$ is the starting height.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle h = \frac{8}{2} = 4$ and $k = 16 - 32 + 3 = -13$.
The parabola opens down with vertex $(-1, 7)$, so the maximum value is 7.
$a = 2$ stretches the graph and $(x - 3)$ moves it right 3.
$\displaystyle x = -\frac{12}{2(2)} = -3$.
It opens up, so the range starts at the vertex value $k = -6$.
The peak is at $\displaystyle t = \frac{80}{32} = 2.5$, and $h(2.5) = -100 + 200 = 100$ feet.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A ball is thrown upward from a platform. Its height in feet after $t$ seconds is $h(t) = -16t^2 + 32t + 48$.
(a) How high is the platform?
Show the answer and scoring guide
At $t = 0$, $h(0) = 48$. The platform is 48 feet high.
Check off the points you earned:
(b) Find the time and height of the ball's maximum point.
Show the answer and scoring guide
$\displaystyle t = -\frac{32}{2(-16)} = 1$ second. $h(1) = -16 + 32 + 48 = 64$ feet.
Check off the points you earned:
(c) When does the ball hit the ground?
Show the answer and scoring guide
Set $-16t^2 + 32t + 48 = 0$. Divide by $-16$: $t^2 - 2t - 3 = 0$, so $(t - 3)(t + 1) = 0$. The answer $t = -1$ makes no sense, so the ball lands at $t = 3$ seconds.
Check off the points you earned: