A y-intercept is a point where a graph crosses the y-axis, meaning x = 0. In AP Precalc Topic 4.2, the y-intercepts of a parametric function f(t) = (x(t), y(t)) occur at the parameter values t where x(t) = 0, because that's when the particle's horizontal position is zero.
A y-intercept is any point where a curve crosses the y-axis. Crossing the y-axis means the x-coordinate is 0, so finding y-intercepts always comes down to one question: when does x equal zero?
In Unit 4, this idea gets a twist. A parametric function f(t) = (x(t), y(t)) describes a particle's position at time t, with separate functions controlling horizontal and vertical position. The particle's path crosses the y-axis whenever its horizontal position is zero, which happens at the real zeros of x(t). That's the flip that trips people up. You solve x(t) = 0, then plug those t-values into y(t) to get the actual intercept points. For example, with f(t) = (t² - 4, t), solving t² - 4 = 0 gives t = ±2, so the y-intercepts are (0, 2) and (0, -2). Notice a parametric path can cross the y-axis more than once, unlike a function of x, which has at most one y-intercept.
This term lives in Topic 4.2: Parametric Functions Modeling Planar Motion (Unit 4) and directly supports learning objective 4.2.A, identifying key characteristics of parametric planar motion related to position. Essential knowledge 4.2.A.3 states it outright: the real zeros of x(t) correspond to y-intercepts, and the real zeros of y(t) correspond to x-intercepts. The exam loves this because it tests whether you actually understand that x(t) and y(t) are two independent functions sharing one input, t. If you can answer "the path crosses the y-axis when x(t) = 0" without hesitation, you've got the core idea of parametric position down. It also pairs with 4.2.A.2, where the max and min values of x(t) and y(t) give the horizontal and vertical extrema of the motion.
Keep studying AP® Precalculus Unit 4
Real zero (Unit 4)
A real zero of a function is an input that makes the output zero. In parametric motion, the real zeros of x(t) are exactly the times the particle sits on the y-axis. Y-intercepts in Topic 4.2 are just real zeros wearing a different outfit.
Zeros of polynomial and rational functions (Unit 1)
The algebra you use here is Unit 1 skill. Solving x(t) = 0 for something like t² - 4 is the same zero-finding you did with polynomials early in the course, just applied to one coordinate of a motion model.
Horizontal and vertical extrema of particle motion (Unit 4)
EK 4.2.A.2 is the sibling idea. Intercepts come from the zeros of x(t) and y(t), while extrema come from their max and min values. Together they map out where a particle's path crosses the axes and how far it travels in each direction.
Expect multiple-choice stems that test the x(t)/y(t) flip directly, like "What do real zeros of x(t) correspond to on the graph?" or "Which characteristic describes the points where the particle's path crosses the y-axis?" The answer in both cases involves x(t), not y(t). You'll also get computational versions, like finding the y-intercepts of f(t) = (t² - 4, t), where you solve x(t) = 0, get t = ±2, and report the points (0, 2) and (0, -2). Two habits keep you safe: always finish by evaluating y(t) at your t-values so you give actual points, not just times, and remember to give the y-intercept as a coordinate pair with x = 0. No released FRQ has centered on this term verbatim, but it's exactly the kind of position characteristic 4.2.A asks you to identify when analyzing parametric motion.
These get crossed constantly in parametric problems because the pairing feels backwards. Y-intercepts come from the zeros of x(t), and x-intercepts come from the zeros of y(t). The logic fixes it: a point on the y-axis has x = 0, so you need the function controlling x to be zero. Whichever axis you're crossing, the OTHER coordinate's function must equal zero.
A y-intercept is a point where the graph crosses the y-axis, which always means the x-coordinate equals 0.
For a parametric function f(t) = (x(t), y(t)), y-intercepts occur at the real zeros of x(t), and x-intercepts occur at the real zeros of y(t), per EK 4.2.A.3.
To find y-intercepts of a parametric path, solve x(t) = 0 for t, then plug each solution into y(t) to get the full point (0, y(t)).
A parametric path can have multiple y-intercepts because the particle can cross the y-axis at several different times.
Don't stop at the t-value. The intercept is a point in the plane, so the exam wants coordinates like (0, 2), not just t = 2.
It's a point where the particle's path crosses the y-axis. For f(t) = (x(t), y(t)), y-intercepts occur at the parameter values where x(t) = 0, which is essential knowledge 4.2.A.3 in Topic 4.2.
No, and this is the classic trap. Zeros of y(t) give x-intercepts, because y = 0 puts the particle on the x-axis. Y-intercepts come from the zeros of x(t), since crossing the y-axis means the horizontal position is zero.
A y-intercept is where the path crosses the y-axis (x = 0, so solve x(t) = 0), while an x-intercept is where it crosses the x-axis (y = 0, so solve y(t) = 0). The function you set to zero is always the one for the coordinate that must vanish.
Set x(t) = t² - 4 equal to 0, giving t = 2 and t = -2. Plug those into y(t) = t to get the y-intercepts (0, 2) and (0, -2).
Yes. Unlike a function of x, which crosses the y-axis at most once, a parametric path crosses the y-axis at every real zero of x(t), so it can hit the y-axis multiple times at different values of t.
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