Apply quadratic functions to real-world optimization problems. Alberta Math 20-1.
Lesson 1.10 of Quadratic Functions & Equations in Math 20-1 — Alberta curriculum lessons.
Application problems look different on the surface but always follow the same algebraic structure. A consistent six-step strategy keeps you organized.
Function vs. Equation in Applications. Before you start, decide which type of problem you have:
Quadratic Function — find a maximum or minimum:
Convert to vertex form y = a(x - p)^2 + q. The vertex gives the max or min.
Quadratic Equation — find when something happens:
Set the function equal to a specific value (often 0) and solve by factoring, completing the square, or the quadratic formula.
What is Projectile Motion?. When something is thrown, kicked, dropped, or shot into the air, its height over time follows a quadratic function. The general form is:
h(t) = -gt^2 + v_0 t + h_0
where g depends on units (gravity), v_0 is initial velocity, and h_0 is initial height.
You do not need to memorize this formula — the function will always be given to you. Focus on reading what the question asks: max height (vertex), landing time (x-intercept), or a specific height (set equal to a value).
Setup — Baseball Height Function. The function h(t) = -6(t - 2.5)^2 + 38.5 gives the height in metres of a batted baseball as a function of time t in seconds.
This is already in vertex form with vertex (2.5, 38.5). Since a = -6 < 0, the parabola opens downward, so the vertex is the maximum point.
Part a) — Maximum Height. The vertex is (2.5, 38.5), so the y-coordinate of the vertex is the maximum.
Answer: Maximum height = 38.5 m
Part b) — Time of Maximum Height. The maximum occurs at the x-coordinate of the vertex.
Answer: t = 2.5 s
Part c) — Height When Hit (at t = 0). Substitute t = 0 into the function:
h(0) = -6(0 - 2.5)^2 + 38.5
= -6(6.25) + 38.5
= -37.5 + 38.5
Answer: h(0) = 1 m
Part e) — Height at t = 2 Seconds. Substitute t = 2:
h(2) = -6(2 - 2.5)^2 + 38.5
= -6(0.25) + 38.5
= -1.5 + 38.5
Answer: h(2) = 37 m
Part f) — When Does the Ball Hit the Ground?. Set h(t) = 0 and solve for t:
Step 1 — Set up:
0 = -6(t - 2.5)^2 + 38.5
Step 2 — Isolate the squared term:
-38.5 = -6(t - 2.5)^2
Step 3 — Divide by -6:
(38.5)/(6) = (t - 2.5)^2
Step 4 — Square root both sides:
±√((38.5)/(6)) = t - 2.5
± 2.53 ≈ t - 2.5
Step 5 — Solve for t:
t ≈ 2.5 + 2.53 = 5.03
t ≈ 2.5 - 2.53 = -0.03 (reject — time cannot be negative)
Answer: The ball hits the ground at t ≈ 5.0 s
Part g) — Domain and Range in Context. Always restrict the domain to the physical situation:
Domain: 0 ≤ t ≤ 5.0 (from when hit until it lands)
Range: 0 ≤ h ≤ 38.5 (from ground to maximum height)
Warning — Always Restrict the Domain in Application Problems. A ball cannot be in the air for negative time or after it has already landed. Always check whether solutions make physical sense and reject any that fall outside the realistic domain.
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